Factoring Trinomials: Leading Coefficient 1 and Not 1

\(\left(x + 3\right) \left(x + 5\right)\)

1. Write the polynomial in standard form and read the coefficients with their signs.

   \[x^{2} + 8 x + 15\]

2. The original signed coefficients are \(a=1,\quad b=8,\quad c=15\). The leading coefficient is 1.

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(15\) (the constant term) and sum \(8\). Use \(3\) and \(5\).

   \[\begin{aligned}(3)(5)&=15\\3+(5)&=8\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 3\right) \left(x + 5\right)\]

\(\left(x + 2\right) \left(2 x + 1\right)\)

1. Write the polynomial in standard form and read the coefficients with their signs.

   \[2 x^{2} + 5 x + 2\]

2. The original signed coefficients are \(a=2,\quad b=5,\quad c=2\). The leading coefficient is not 1.

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(4\) (a × c) and sum \(5\). Use \(1\) and \(4\).

   \[\begin{aligned}(1)(4)&=4\\1+(4)&=5\end{aligned}\]

5. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} + x + 4 x + 2\]

6. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(2 x + 1\right) + 2\left(2 x + 1\right)\]

7. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 2\right) \left(2 x + 1\right)\]

\(\left(x - 2\right) \left(4 x - 1\right)\)

1. Write the polynomial in standard form and read the coefficients with their signs.

   \[4 x^{2} - 9 x + 2\]

2. The original signed coefficients are \(a=4,\quad b=-9,\quad c=2\). The leading coefficient is not 1.

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(8\) (a × c) and sum \(-9\). Use \(-8\) and \(-1\).

   \[\begin{aligned}(-8)(-1)&=8\\-8+(-1)&=-9\end{aligned}\]

5. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[4 x^{2} - 8 x - x + 2\]

6. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[4 x\left(x - 2\right) - 1\left(x - 2\right)\]

7. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x - 2\right) \left(4 x - 1\right)\]

\(\left(x + 3\right) \left(x + 8\right)\)

1. Write the polynomial in standard form and read the coefficients with their signs.

   \[x^{2} + 11 x + 24\]

2. The original signed coefficients are \(a=1,\quad b=11,\quad c=24\). The leading coefficient is 1.

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(24\) (the constant term) and sum \(11\). Use \(3\) and \(8\).

   \[\begin{aligned}(3)(8)&=24\\3+(8)&=11\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 3\right) \left(x + 8\right)\]

\(\left(4 - 3 x\right) \left(x - 2\right)\)

1. Write the polynomial in standard form and read the coefficients with their signs.

   \[- 3 x^{2} + 10 x - 8\]

2. The original signed coefficients are \(a=-3,\quad b=10,\quad c=-8\). The leading coefficient is not 1.

3. Factor out \(-1\). Divide every term by this common factor and keep it outside.

   \[-1\left(3 x^{2} - 10 x + 8\right)\]

4. Find two integers with product \(24\) (a × c) and sum \(-10\). Use \(-6\) and \(-4\).

   \[\begin{aligned}(-6)(-4)&=24\\-6+(-4)&=-10\end{aligned}\]

5. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[3 x^{2} - 6 x - 4 x + 8\]

6. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[3 x\left(x - 2\right) - 4\left(x - 2\right)\]

7. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(4 - 3 x\right) \left(x - 2\right)\]

\(\left(x + 5\right) \left(4 x + 8\right)\)

1. Factor out \(4\). Divide every term by this common factor and keep it outside.

   \[4\left(x^{2} + 7 x + 10\right)\]

2. Find two integers with product \(10\) (the constant term) and sum \(7\). Use \(2\) and \(5\).

   \[\begin{aligned}(2)(5)&=10\\2+(5)&=7\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 5\right) \left(4 x + 8\right)\]

\(\left(x - 5\right) \left(6 x - 12\right)\)

1. Factor out \(6\). Divide every term by this common factor and keep it outside.

   \[6\left(x^{2} - 7 x + 10\right)\]

2. Find two integers with product \(10\) (the constant term) and sum \(-7\). Use \(-5\) and \(-2\).

   \[\begin{aligned}(-5)(-2)&=10\\-5+(-2)&=-7\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 5\right) \left(6 x - 12\right)\]

\(3 x \left(x + 3\right) \left(x + 5\right)\)

1. Factor out \(3 x\). Divide every term by this common factor and keep it outside.

   \[3 x\left(x^{2} + 8 x + 15\right)\]

2. Find two integers with product \(15\) (the constant term) and sum \(8\). Use \(3\) and \(5\).

   \[\begin{aligned}(3)(5)&=15\\3+(5)&=8\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[3 x \left(x + 3\right) \left(x + 5\right)\]

\(\left(- 2 x - 8\right) \left(x + 5\right)\)

1. Factor out \(-2\). Divide every term by this common factor and keep it outside.

   \[-2\left(x^{2} + 9 x + 20\right)\]

2. Find two integers with product \(20\) (the constant term) and sum \(9\). Use \(4\) and \(5\).

   \[\begin{aligned}(4)(5)&=20\\4+(5)&=9\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(- 2 x - 8\right) \left(x + 5\right)\]

\(6 x^{2} \left(x - 2\right) \left(2 x - 1\right)\)

1. Factor out \(6 x^{2}\). Divide every term by this common factor and keep it outside.

   \[6 x^{2}\left(2 x^{2} - 5 x + 2\right)\]

2. Find two integers with product \(4\) (a × c) and sum \(-5\). Use \(-4\) and \(-1\).

   \[\begin{aligned}(-4)(-1)&=4\\-4+(-1)&=-5\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} - 4 x - x + 2\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(x - 2\right) - 1\left(x - 2\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[6 x^{2} \left(x - 2\right) \left(2 x - 1\right)\]

\(\left(x + 3\right) \left(x + 7\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(21\) (the constant term) and sum \(10\). Use \(3\) and \(7\).

   \[\begin{aligned}(3)(7)&=21\\3+(7)&=10\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 3\right) \left(x + 7\right)\]

\(\left(x + 5\right) \left(x + 7\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(35\) (the constant term) and sum \(12\). Use \(5\) and \(7\).

   \[\begin{aligned}(5)(7)&=35\\5+(7)&=12\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 5\right) \left(x + 7\right)\]

\(\left(x - 7\right) \left(x - 4\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(28\) (the constant term) and sum \(-11\). Use \(-7\) and \(-4\).

   \[\begin{aligned}(-7)(-4)&=28\\-7+(-4)&=-11\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 7\right) \left(x - 4\right)\]

\(\left(x - 9\right) \left(x - 8\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(72\) (the constant term) and sum \(-17\). Use \(-9\) and \(-8\).

   \[\begin{aligned}(-9)(-8)&=72\\-9+(-8)&=-17\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 9\right) \left(x - 8\right)\]

\(\left(x - 14\right) \left(x - 9\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(126\) (the constant term) and sum \(-23\). Use \(-14\) and \(-9\).

   \[\begin{aligned}(-14)(-9)&=126\\-14+(-9)&=-23\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 14\right) \left(x - 9\right)\]

\(\left(x - 3\right) \left(x + 8\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-24\) (the constant term) and sum \(5\). Use \(-3\) and \(8\).

   \[\begin{aligned}(-3)(8)&=-24\\-3+(8)&=5\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 3\right) \left(x + 8\right)\]

\(\left(x - 9\right) \left(x + 4\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-36\) (the constant term) and sum \(-5\). Use \(-9\) and \(4\).

   \[\begin{aligned}(-9)(4)&=-36\\-9+(4)&=-5\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 9\right) \left(x + 4\right)\]

\(\left(x - 7\right) \left(x + 9\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-63\) (the constant term) and sum \(2\). Use \(-7\) and \(9\).

   \[\begin{aligned}(-7)(9)&=-63\\-7+(9)&=2\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 7\right) \left(x + 9\right)\]

\(\left(x - 11\right) \left(x + 7\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-77\) (the constant term) and sum \(-4\). Use \(-11\) and \(7\).

   \[\begin{aligned}(-11)(7)&=-77\\-11+(7)&=-4\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 11\right) \left(x + 7\right)\]

\(\left(x - 14\right) \left(x + 11\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-154\) (the constant term) and sum \(-3\). Use \(-14\) and \(11\).

   \[\begin{aligned}(-14)(11)&=-154\\-14+(11)&=-3\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 14\right) \left(x + 11\right)\]

\(\left(x + 4\right) \left(2 x + 1\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(8\) (a × c) and sum \(9\). Use \(1\) and \(8\).

   \[\begin{aligned}(1)(8)&=8\\1+(8)&=9\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} + x + 8 x + 4\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(2 x + 1\right) + 4\left(2 x + 1\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 4\right) \left(2 x + 1\right)\]

\(\left(x + 3\right) \left(3 x + 2\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(18\) (a × c) and sum \(11\). Use \(2\) and \(9\).

   \[\begin{aligned}(2)(9)&=18\\2+(9)&=11\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[3 x^{2} + 2 x + 9 x + 6\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(3 x + 2\right) + 3\left(3 x + 2\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 3\right) \left(3 x + 2\right)\]

\(\left(x + 2\right) \left(4 x + 3\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(24\) (a × c) and sum \(11\). Use \(3\) and \(8\).

   \[\begin{aligned}(3)(8)&=24\\3+(8)&=11\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[4 x^{2} + 3 x + 8 x + 6\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(4 x + 3\right) + 2\left(4 x + 3\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 2\right) \left(4 x + 3\right)\]

\(\left(x + 3\right) \left(5 x + 2\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(30\) (a × c) and sum \(17\). Use \(2\) and \(15\).

   \[\begin{aligned}(2)(15)&=30\\2+(15)&=17\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[5 x^{2} + 2 x + 15 x + 6\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(5 x + 2\right) + 3\left(5 x + 2\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 3\right) \left(5 x + 2\right)\]

\(\left(2 x + 5\right) \left(3 x + 2\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(60\) (a × c) and sum \(19\). Use \(4\) and \(15\).

   \[\begin{aligned}(4)(15)&=60\\4+(15)&=19\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[6 x^{2} + 4 x + 15 x + 10\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(3 x + 2\right) + 5\left(3 x + 2\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(2 x + 5\right) \left(3 x + 2\right)\]

\(\left(x + 4\right) \left(2 x - 1\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-8\) (a × c) and sum \(7\). Use \(-1\) and \(8\).

   \[\begin{aligned}(-1)(8)&=-8\\-1+(8)&=7\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} - x + 8 x - 4\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(2 x - 1\right) + 4\left(2 x - 1\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 4\right) \left(2 x - 1\right)\]

\(\left(x - 3\right) \left(3 x - 1\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(9\) (a × c) and sum \(-10\). Use \(-9\) and \(-1\).

   \[\begin{aligned}(-9)(-1)&=9\\-9+(-1)&=-10\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[3 x^{2} - 9 x - x + 3\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[3 x\left(x - 3\right) - 1\left(x - 3\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x - 3\right) \left(3 x - 1\right)\]

\(\left(2 x - 3\right) \left(2 x + 5\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-60\) (a × c) and sum \(4\). Use \(-6\) and \(10\).

   \[\begin{aligned}(-6)(10)&=-60\\-6+(10)&=4\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[4 x^{2} - 6 x + 10 x - 15\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(2 x - 3\right) + 5\left(2 x - 3\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(2 x - 3\right) \left(2 x + 5\right)\]

\(\left(2 x - 3\right) \left(3 x - 2\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(36\) (a × c) and sum \(-13\). Use \(-9\) and \(-4\).

   \[\begin{aligned}(-9)(-4)&=36\\-9+(-4)&=-13\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[6 x^{2} - 9 x - 4 x + 6\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[3 x\left(2 x - 3\right) - 2\left(2 x - 3\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(2 x - 3\right) \left(3 x - 2\right)\]

\(\left(2 x + 1\right) \left(4 x - 3\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-24\) (a × c) and sum \(-2\). Use \(-6\) and \(4\).

   \[\begin{aligned}(-6)(4)&=-24\\-6+(4)&=-2\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[8 x^{2} - 6 x + 4 x - 3\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(4 x - 3\right) + 1\left(4 x - 3\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(2 x + 1\right) \left(4 x - 3\right)\]

\(\left(x + 6\right)^{2}\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(36\) (the constant term) and sum \(12\). Use \(6\) and \(6\).

   \[\begin{aligned}(6)(6)&=36\\6+(6)&=12\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 6\right)^{2}\]

4. The repeated binomial can be written as a square. Its middle term is twice the product of its two terms.

   \[2\left(x\right)\left(6\right) = 12 x\]

\(\left(3 x - 5\right)^{2}\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(225\) (a × c) and sum \(-30\). Use \(-15\) and \(-15\).

   \[\begin{aligned}(-15)(-15)&=225\\-15+(-15)&=-30\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[9 x^{2} - 15 x - 15 x + 25\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[3 x\left(3 x - 5\right) - 5\left(3 x - 5\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(3 x - 5\right)^{2}\]

6. The repeated binomial can be written as a square. Its middle term is twice the product of its two terms.

   \[2\left(3 x\right)\left(-5\right) = - 30 x\]

\(3 \left(x - 4\right)^{2}\)

1. Factor out \(3\). Divide every term by this common factor and keep it outside.

   \[3\left(x^{2} - 8 x + 16\right)\]

2. Find two integers with product \(16\) (the constant term) and sum \(-8\). Use \(-4\) and \(-4\).

   \[\begin{aligned}(-4)(-4)&=16\\-4+(-4)&=-8\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[3 \left(x - 4\right)^{2}\]

4. The repeated binomial can be written as a square. Its middle term is twice the product of its two terms.

   \[2\left(x\right)\left(-4\right) = - 8 x\]

Irreducible over the integers.

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. For an integer factorization, the middle-term pair must have product -4 and sum 1.

3. All distinct signed integer pairs with that product are: -4 and 1 (sum -3); -1 and 4 (sum 3); -2 and 2 (sum 0).

4. None has the required sum. Therefore the quadratic is irreducible over the integers; this statement is about the allowed coefficients, not a claim about every larger number system.

Irreducible over the integers.

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. For an integer factorization, the middle-term pair must have product 6 and sum 3.

3. All distinct signed integer pairs with that product are: 1 and 6 (sum 7); -6 and -1 (sum -7); 2 and 3 (sum 5); -3 and -2 (sum -5).

4. None has the required sum. Therefore the quadratic is irreducible over the integers; this statement is about the allowed coefficients, not a claim about every larger number system.

\(\left(x + 4\right) \left(3 x + 2\right)\)

1. The proposed product expands to \(3x^2+10x+8\). Its middle coefficient is 10, not 14, so it does not match the original.

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(24\) (a × c) and sum \(14\). Use \(2\) and \(12\).

   \[\begin{aligned}(2)(12)&=24\\2+(12)&=14\end{aligned}\]

4. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[3 x^{2} + 2 x + 12 x + 8\]

5. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(3 x + 2\right) + 4\left(3 x + 2\right)\]

6. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 4\right) \left(3 x + 2\right)\]

\(x=3\ \text{or}\ x=5\)

1. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[x^{2} - 8 x + 15 = 0\]

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(15\) (the constant term) and sum \(-8\). Use \(-5\) and \(-3\).

   \[\begin{aligned}(-5)(-3)&=15\\-5+(-3)&=-8\end{aligned}\]

4. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 5\right) \left(x - 3\right)\]

5. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 5\right) \left(x - 3\right) = 0\]

6. Solve this factor equation.

   \[x - 5=0\quad\Longrightarrow\quad x=5\]

7. Solve this factor equation.

   \[x - 3=0\quad\Longrightarrow\quad x=3\]

8. Collect the algebraic solutions.

   \[x=3\ \text{or}\ x=5\]

\(x=-5\ \text{or}\ x=\frac{3}{2}\)

1. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[2 x^{2} + 7 x - 15 = 0\]

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(-30\) (a × c) and sum \(7\). Use \(-3\) and \(10\).

   \[\begin{aligned}(-3)(10)&=-30\\-3+(10)&=7\end{aligned}\]

4. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} - 3 x + 10 x - 15\]

5. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(2 x - 3\right) + 5\left(2 x - 3\right)\]

6. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 5\right) \left(2 x - 3\right)\]

7. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x + 5\right) \left(2 x - 3\right) = 0\]

8. Solve this factor equation.

   \[x + 5=0\quad\Longrightarrow\quad x=-5\]

9. Solve this factor equation.

   \[2 x - 3=0\quad\Longrightarrow\quad x=\frac{3}{2}\]

10. Collect the algebraic solutions.

    \[x=-5\ \text{or}\ x=\frac{3}{2}\]

\(x=\frac{2}{3}\ \text{or}\ x=4\)

1. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[3 x^{2} - 14 x + 8 = 0\]

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(24\) (a × c) and sum \(-14\). Use \(-12\) and \(-2\).

   \[\begin{aligned}(-12)(-2)&=24\\-12+(-2)&=-14\end{aligned}\]

4. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[3 x^{2} - 12 x - 2 x + 8\]

5. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[3 x\left(x - 4\right) - 2\left(x - 4\right)\]

6. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x - 4\right) \left(3 x - 2\right)\]

7. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 4\right) \left(3 x - 2\right) = 0\]

8. Solve this factor equation.

   \[x - 4=0\quad\Longrightarrow\quad x=4\]

9. Solve this factor equation.

   \[3 x - 2=0\quad\Longrightarrow\quad x=\frac{2}{3}\]

10. Collect the algebraic solutions.

    \[x=\frac{2}{3}\ \text{or}\ x=4\]

\(x=0\ \text{or}\ x=2\ \text{or}\ x=5\)

1. Do not divide both sides by \(x\): that would assume \(x\ne0\) and could discard the solution \(x=0\). Factor the variable out instead.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[2 x^{3} - 14 x^{2} + 20 x = 0\]

3. Factor out \(2 x\). Divide every term by this common factor and keep it outside.

   \[2 x\left(x^{2} - 7 x + 10\right)\]

4. Find two integers with product \(10\) (the constant term) and sum \(-7\). Use \(-5\) and \(-2\).

   \[\begin{aligned}(-5)(-2)&=10\\-5+(-2)&=-7\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[2 x \left(x - 5\right) \left(x - 2\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[2 x \left(x - 5\right) \left(x - 2\right) = 0\]

7. Solve this factor equation.

   \[x - 5=0\quad\Longrightarrow\quad x=5\]

8. Solve this factor equation.

   \[x - 2=0\quad\Longrightarrow\quad x=2\]

9. Solve this factor equation.

   \[x=0\quad\Longrightarrow\quad x=0\]

10. Collect the algebraic solutions.

    \[x=0\ \text{or}\ x=2\ \text{or}\ x=5\]

\(\left(x - 3\right) \left(5 x - 4\right)\)

1. Write the polynomial in standard form and read the coefficients with their signs.

   \[5 x^{2} - 19 x + 12\]

2. The original signed coefficients are \(a=5,\quad b=-19,\quad c=12\). The leading coefficient is not 1.

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(60\) (a × c) and sum \(-19\). Use \(-15\) and \(-4\).

   \[\begin{aligned}(-15)(-4)&=60\\-15+(-4)&=-19\end{aligned}\]

5. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[5 x^{2} - 15 x - 4 x + 12\]

6. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[5 x\left(x - 3\right) - 4\left(x - 3\right)\]

7. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x - 3\right) \left(5 x - 4\right)\]

\(\left(x + 5\right) \left(6 x + 24\right)\)

1. Factor out \(6\). Divide every term by this common factor and keep it outside.

   \[6\left(x^{2} + 9 x + 20\right)\]

2. Find two integers with product \(20\) (the constant term) and sum \(9\). Use \(4\) and \(5\).

   \[\begin{aligned}(4)(5)&=20\\4+(5)&=9\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 5\right) \left(6 x + 24\right)\]

\(\left(x - 9\right) \left(x - 7\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(63\) (the constant term) and sum \(-16\). Use \(-9\) and \(-7\).

   \[\begin{aligned}(-9)(-7)&=63\\-9+(-7)&=-16\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 9\right) \left(x - 7\right)\]

\(\left(x - 5\right) \left(x + 9\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-45\) (the constant term) and sum \(4\). Use \(-5\) and \(9\).

   \[\begin{aligned}(-5)(9)&=-45\\-5+(9)&=4\end{aligned}\]

3. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 5\right) \left(x + 9\right)\]

\(\left(x + 4\right) \left(3 x + 1\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(12\) (a × c) and sum \(13\). Use \(1\) and \(12\).

   \[\begin{aligned}(1)(12)&=12\\1+(12)&=13\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[3 x^{2} + x + 12 x + 4\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(3 x + 1\right) + 4\left(3 x + 1\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 4\right) \left(3 x + 1\right)\]

\(\left(2 x + 1\right) \left(5 x - 3\right)\)

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. Find two integers with product \(-30\) (a × c) and sum \(-1\). Use \(-6\) and \(5\).

   \[\begin{aligned}(-6)(5)&=-30\\-6+(5)&=-1\end{aligned}\]

3. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[10 x^{2} - 6 x + 5 x - 3\]

4. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(5 x - 3\right) + 1\left(5 x - 3\right)\]

5. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(2 x + 1\right) \left(5 x - 3\right)\]

Irreducible over the integers.

1. The GCF is 1, so no common numerical or variable factor needs to be removed.

2. For an integer factorization, the middle-term pair must have product 7 and sum 4.

3. All distinct signed integer pairs with that product are: 1 and 7 (sum 8); -7 and -1 (sum -8).

4. None has the required sum. Therefore the quadratic is irreducible over the integers; this statement is about the allowed coefficients, not a claim about every larger number system.

\(x=- \frac{3}{2}\ \text{or}\ x=\frac{5}{2}\)

1. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[4 x^{2} - 4 x - 15 = 0\]

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(-60\) (a × c) and sum \(-4\). Use \(-10\) and \(6\).

   \[\begin{aligned}(-10)(6)&=-60\\-10+(6)&=-4\end{aligned}\]

4. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[4 x^{2} - 10 x + 6 x - 15\]

5. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(2 x - 5\right) + 3\left(2 x - 5\right)\]

6. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(2 x - 5\right) \left(2 x + 3\right)\]

7. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(2 x - 5\right) \left(2 x + 3\right) = 0\]

8. Solve this factor equation.

   \[2 x - 5=0\quad\Longrightarrow\quad x=\frac{5}{2}\]

9. Solve this factor equation.

   \[2 x + 3=0\quad\Longrightarrow\quad x=- \frac{3}{2}\]

10. Collect the algebraic solutions.

    \[x=- \frac{3}{2}\ \text{or}\ x=\frac{5}{2}\]

\(\left(x + 3\right) \left(x + 6\right)\)

1. Area equals the product of the two side lengths. Match the known side to one factor.

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(18\) (the constant term) and sum \(9\). Use \(3\) and \(6\).

   \[\begin{aligned}(3)(6)&=18\\3+(6)&=9\end{aligned}\]

4. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 3\right) \left(x + 6\right)\]

In context: The other side is x + 6 cm. At x = 4, the sides are 7 cm and 10 cm, giving area 70 cm².

\(\left(x - 7\right) \left(x - 3\right)\)

1. The restriction x > 7 ensures both proposed side lengths are positive.

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(21\) (the constant term) and sum \(-10\). Use \(-7\) and \(-3\).

   \[\begin{aligned}(-7)(-3)&=21\\-7+(-3)&=-10\end{aligned}\]

4. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 7\right) \left(x - 3\right)\]

In context: The other side is x − 7 units. At x = 10, the sides are 7 and 3 units, and the area is 21 square units.

\(\left(x + 4\right) \left(2 x + 3\right)\)

1. Factor the area and identify the factor paired with the known width.

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(24\) (a × c) and sum \(11\). Use \(3\) and \(8\).

   \[\begin{aligned}(3)(8)&=24\\3+(8)&=11\end{aligned}\]

4. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} + 3 x + 8 x + 12\]

5. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[x\left(2 x + 3\right) + 4\left(2 x + 3\right)\]

6. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x + 4\right) \left(2 x + 3\right)\]

In context: The length is x + 4 feet. At x = 2, the dimensions are 7 feet by 6 feet and the area is 42 square feet.

\(\left(x + 9\right)^{2}\)

1. Recognize a perfect square. A physical side length must be positive.

2. The GCF is 1, so no common numerical or variable factor needs to be removed.

3. Find two integers with product \(81\) (the constant term) and sum \(18\). Use \(9\) and \(9\).

   \[\begin{aligned}(9)(9)&=81\\9+(9)&=18\end{aligned}\]

4. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 9\right)^{2}\]

5. The repeated binomial can be written as a square. Its middle term is twice the product of its two terms.

   \[2\left(x\right)\left(9\right) = 18 x\]

In context: The side length is x + 9 cm. At x = 3 it is 12 cm, so the perimeter is 4 × 12 = 48 cm.

\(\left(x + 5\right) \left(3 x + 6\right)\)

1. The common factor 3 represents the three identical banners. The remaining quadratic is the area of one banner.

2. Factor out \(3\). Divide every term by this common factor and keep it outside.

   \[3\left(x^{2} + 7 x + 10\right)\]

3. Find two integers with product \(10\) (the constant term) and sum \(7\). Use \(2\) and \(5\).

   \[\begin{aligned}(2)(5)&=10\\2+(5)&=7\end{aligned}\]

4. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x + 5\right) \left(3 x + 6\right)\]

In context: Each banner has length x + 5 feet. At x = 4, one banner is 6 by 9 feet and has area 54 square feet. The three banners have total area 162 square feet.

\(x=-12\ \text{or}\ x=7\)

1. Let x be the width in meters, with x > 0. The length is x + 5.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[x^{2} + 5 x - 84 = 0\]

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(-84\) (the constant term) and sum \(5\). Use \(-7\) and \(12\).

   \[\begin{aligned}(-7)(12)&=-84\\-7+(12)&=5\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 7\right) \left(x + 12\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 7\right) \left(x + 12\right) = 0\]

7. Solve this factor equation.

   \[x - 7=0\quad\Longrightarrow\quad x=7\]

8. Solve this factor equation.

   \[x + 12=0\quad\Longrightarrow\quad x=-12\]

9. Collect the algebraic solutions.

   \[x=-12\ \text{or}\ x=7\]

In context: The roots are −12 and 7. Reject the negative width; the garden is 7 meters wide and 12 meters long, and 7 × 12 = 84.

\(x=-12\ \text{or}\ x=11\)

1. Let x be the smaller positive integer. The next integer is x + 1.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[x^{2} + x - 132 = 0\]

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(-132\) (the constant term) and sum \(1\). Use \(-11\) and \(12\).

   \[\begin{aligned}(-11)(12)&=-132\\-11+(12)&=1\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 11\right) \left(x + 12\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 11\right) \left(x + 12\right) = 0\]

7. Solve this factor equation.

   \[x - 11=0\quad\Longrightarrow\quad x=11\]

8. Solve this factor equation.

   \[x + 12=0\quad\Longrightarrow\quad x=-12\]

9. Collect the algebraic solutions.

   \[x=-12\ \text{or}\ x=11\]

In context: Reject x = −12 because the integers must be positive. The integers are 11 and 12, and 11 × 12 = 132.

\(x=- \frac{13}{2}\ \text{or}\ x=5\)

1. Let x be the positive width in centimeters. The length is 2x + 3.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[2 x^{2} + 3 x - 65 = 0\]

3. The GCF is 1, so no common numerical or variable factor needs to be removed.

4. Find two integers with product \(-130\) (a × c) and sum \(3\). Use \(-10\) and \(13\).

   \[\begin{aligned}(-10)(13)&=-130\\-10+(13)&=3\end{aligned}\]

5. Split the middle term into two terms with the selected coefficients. The following work is for the quadratic inside any outside GCF.

   \[2 x^{2} - 10 x + 13 x - 65\]

6. Factor each group. The binomial inside both groups must match; factor a negative coefficient when needed.

   \[2 x\left(x - 5\right) + 13\left(x - 5\right)\]

7. Factor out the common binomial and restore the outside GCF, if there was one.

   \[\left(x - 5\right) \left(2 x + 13\right)\]

8. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 5\right) \left(2 x + 13\right) = 0\]

9. Solve this factor equation.

   \[x - 5=0\quad\Longrightarrow\quad x=5\]

10. Solve this factor equation.

    \[2 x + 13=0\quad\Longrightarrow\quad x=- \frac{13}{2}\]

11. Collect the algebraic solutions.

    \[x=- \frac{13}{2}\ \text{or}\ x=5\]

In context: The roots are −13/2 and 5. Only the positive width is valid: the sign is 5 cm wide and 13 cm long, and 5 × 13 = 65.

\(x=-9\ \text{or}\ x=1\)

1. The outside dimensions are 6 + 2x and 10 + 2x because the path adds width on BOTH sides. A physical path requires x > 0.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[4 x^{2} + 32 x - 36 = 0\]

3. Factor out \(4\). Divide every term by this common factor and keep it outside.

   \[4\left(x^{2} + 8 x - 9\right)\]

4. Find two integers with product \(-9\) (the constant term) and sum \(8\). Use \(-1\) and \(9\).

   \[\begin{aligned}(-1)(9)&=-9\\-1+(9)&=8\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 1\right) \left(4 x + 36\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 1\right) \left(4 x + 36\right) = 0\]

7. Solve this factor equation.

   \[x - 1=0\quad\Longrightarrow\quad x=1\]

8. Solve this factor equation.

   \[x + 9=0\quad\Longrightarrow\quad x=-9\]

9. Collect the algebraic solutions.

   \[x=-9\ \text{or}\ x=1\]

In context: Reject x = −9. The path is 1 meter wide, making the outside dimensions 8 m by 12 m. Their product is 96 m².

\(x=10\ \text{or}\ x=20\)

1. Revenue equals quantity times price. Require x to be an integer with 0 < x < 30.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[- x^{2} + 30 x - 200 = 0\]

3. Factor out \(-1\). Divide every term by this common factor and keep it outside.

   \[-1\left(x^{2} - 30 x + 200\right)\]

4. Find two integers with product \(200\) (the constant term) and sum \(-30\). Use \(-20\) and \(-10\).

   \[\begin{aligned}(-20)(-10)&=200\\-20+(-10)&=-30\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(10 - x\right) \left(x - 20\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(10 - x\right) \left(x - 20\right) = 0\]

7. Solve this factor equation.

   \[x - 20=0\quad\Longrightarrow\quad x=20\]

8. Solve this factor equation.

   \[x - 10=0\quad\Longrightarrow\quad x=10\]

9. Collect the algebraic solutions.

   \[x=10\ \text{or}\ x=20\]

In context: Both options are valid: 10 bags at $20 each or 20 bags at $10 each. Each produces $200 in revenue.

\(x=-1\ \text{or}\ x=4\)

1. Ground level means height zero. Set h(x) = 0 and require elapsed time x ≥ 0.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[- 5 x^{2} + 15 x + 20 = 0\]

3. Factor out \(-5\). Divide every term by this common factor and keep it outside.

   \[-5\left(x^{2} - 3 x - 4\right)\]

4. Find two integers with product \(-4\) (the constant term) and sum \(-3\). Use \(-4\) and \(1\).

   \[\begin{aligned}(-4)(1)&=-4\\-4+(1)&=-3\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(20 - 5 x\right) \left(x + 1\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(20 - 5 x\right) \left(x + 1\right) = 0\]

7. Solve this factor equation.

   \[x - 4=0\quad\Longrightarrow\quad x=4\]

8. Solve this factor equation.

   \[x + 1=0\quad\Longrightarrow\quad x=-1\]

9. Collect the algebraic solutions.

   \[x=-1\ \text{or}\ x=4\]

In context: The roots are −1 and 4. Discard the negative time. The ball reaches the ground after 4 seconds in this model; −5(4²) + 15(4) + 20 = 0.

\(x=3\ \text{or}\ x=19\)

1. The base dimensions are 18 − 2x and 26 − 2x. Require 0 < x < 9 so both base dimensions and the height are positive.

2. Subtract the entire right side from both sides, expand as needed, and combine terms so one side is zero.

   \[4 x^{2} - 88 x + 228 = 0\]

3. Factor out \(4\). Divide every term by this common factor and keep it outside.

   \[4\left(x^{2} - 22 x + 57\right)\]

4. Find two integers with product \(57\) (the constant term) and sum \(-22\). Use \(-19\) and \(-3\).

   \[\begin{aligned}(-19)(-3)&=57\\-19+(-3)&=-22\end{aligned}\]

5. Use the pair as the constants in the two binomial factors, retaining any outside GCF.

   \[\left(x - 19\right) \left(4 x - 12\right)\]

6. The factored product is zero. A nonzero numerical factor cannot make it zero, so set each variable-containing factor equal to zero.

   \[\left(x - 19\right) \left(4 x - 12\right) = 0\]

7. Solve this factor equation.

   \[x - 19=0\quad\Longrightarrow\quad x=19\]

8. Solve this factor equation.

   \[x - 3=0\quad\Longrightarrow\quad x=3\]

9. Collect the algebraic solutions.

   \[x=3\ \text{or}\ x=19\]

In context: The roots are 3 and 19. Only x = 3 lies between 0 and 9. The base is 12 cm by 20 cm and the height is 3 cm, giving volume 12 × 20 × 3 = 720 cm³.