What does it mean to factor a polynomial?

Factoring a polynomial means rewriting it as a product of simpler polynomials. For example, x^2 + 5x + 6 factors as (x + 2)(x + 3). If you multiply the factors back together you should get the original polynomial. Factoring is the inverse of multiplying out.

What is the Remainder Theorem?

The Remainder Theorem says: when a polynomial f(x) is divided by (x − a), the remainder is f(a). So instead of doing long division, you just evaluate the polynomial at x = a. If f(x) = x^3 + 2x^2 − 5x + 3 and we divide by x − 2, the remainder is f(2) = 9.

What is the Factor Theorem?

The Factor Theorem is the special case of the Remainder Theorem where the remainder is zero. It says: (x − a) is a factor of f(x) if and only if f(a) = 0. To test whether (x − 2) is a factor of f(x), just calculate f(2) — if it is zero, then (x − 2) is a factor.

What is the sign trap in the Factor Theorem?

The Factor Theorem is about factors of the form (x − a). To test a factor like (x + 3), rewrite it as (x − (−3)) and test f(−3). Rule of thumb: for factor (x − a) test f(a); for factor (x + a) test f(−a). Always use the OPPOSITE of the number you see in the factor.

How do you completely factor a cubic polynomial?

First find one factor by trying small values of a — the Factor Theorem says (x − a) is a factor if f(a) = 0. Divide the cubic by (x − a) using synthetic division; the quotient is a quadratic. Factor that quadratic by the usual methods. The zeros of the polynomial are the values that make each factor zero.

How are factors, zeros, and x-intercepts of a polynomial related?

They are three ways of describing the same thing. If (x − a) is a factor of f(x), then x = a is a zero (or root) of the polynomial, and the point (a, 0) is an x-intercept of the graph. Polynomial → factor → zero → x-intercept is the central chain of ideas in Algebra II.

Young Fermats — Algebra II · Class 11 of 24 · Grades 7–10

Algebra II Class 11: Factoring Polynomials & the Remainder and Factor Theorems

By the end of this class an honors Algebra II student can factor a polynomial using the five standard techniques, use the Remainder Theorem to find a remainder without doing long division, use the Factor Theorem to test whether a linear expression is a factor of a polynomial, and completely factor a cubic by pairing the Factor Theorem with synthetic division. Below you’ll find the concept map for the class, seven worked examples, and 25 practice questions with click-to-reveal step-by-step answers — plus a matching slide pack and a print worksheet.

Direct answer: Factoring rewrites a polynomial as a product of simpler polynomials. The Remainder Theorem says the remainder when \(f(x)\) is divided by \((x - a)\) is just \(f(a)\). The Factor Theorem says \((x - a)\) is a factor of \(f(x)\) if and only if \(f(a) = 0\). Together they turn hard division problems into simple substitution — and give you the fastest way to find the zeros of a cubic or quartic.

Class 11 slide pack — 12 slides

Every board slide used in class: intro to factoring, GCF, trinomials, difference of squares, perfect square trinomials, grouping, the zeros connection, the Remainder Theorem, the Factor Theorem, the sign trap, completely factoring a cubic, and the Algebra II big-picture chain.

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What’s in this class

What factoring actually means
The five-technique factoring toolbox
Factors, zeros, and \(x\)-intercepts
The Remainder Theorem
The Factor Theorem
The sign trap in \(x + a\) factors
Completely factoring a cubic (Factor Theorem + synthetic division)
The Algebra II big picture
Practice — 25 questions with hidden answers
FAQ for parents
What’s next

1. What factoring actually means

Factoring means rewriting a polynomial as a product of simpler polynomials. It is the exact inverse of “multiplying out.” When we multiplied \((x+2)(x+3)\) in earlier classes we got \(x^2 + 5x + 6\). Factoring is going the other direction: starting from \(x^2 + 5x + 6\) and getting back to \((x+2)(x+3)\).

The check is always the same: if you multiply your factored answer back out and you don’t recover the original polynomial, you factored it wrong. This one-line quality check catches almost every mistake.

Why we care. Factoring lets us simplify rational expressions, solve polynomial equations, find zeros of functions, and describe where a graph crosses the \(x\)-axis. Every one of those uses shows up on the Algebra II Regents, SAT, and Precalculus placement tests.

2. The five-technique factoring toolbox

Ninety-five percent of the factoring problems a student sees before Precalculus fall into one of five patterns. When we hand a student a polynomial in class, the first thing we ask them out loud is: “Which of the five tools does this look like?”

1. Greatest common factor (GCF)

Always look first. Pull out the largest number and the largest variable power that divides every term.

\(6x^3 + 9x^2 = 3x^2(2x + 3)\)

2. Trinomial \(x^2 + bx + c\)

Find two numbers that multiply to \(c\) and add to \(b\).

\(x^2 + 5x + 6 = (x+2)(x+3)\)

3. Difference of squares

Whenever you see \(a^2 - b^2\), factor as \((a-b)(a+b)\). Sum of squares does not factor over the reals.

\(x^2 - 9 = (x - 3)(x + 3)\)

4. Perfect square trinomial

If \(a^2 \pm 2ab + b^2\), it factors as \((a \pm b)^2\). The middle term is the giveaway: it’s twice the product of the square roots of the outer terms.

\(x^2 + 6x + 9 = (x + 3)^2\)

5. Grouping (four terms)

Group into two pairs, factor a common piece from each pair, then factor the common binomial.

\(x^3 + 2x^2 + 3x + 6 = (x+2)(x^2 + 3)\)

Master order

1) GCF first — always. 2) Count terms. Two terms → difference of squares. Three terms → trinomial or perfect square. Four terms → grouping. 3) Check your factors are themselves completely factored.

GCF → count → check.

Worked example — combining tools. Factor \(2x^3 - 8x\) completely.

GCF first: \(2x^3 - 8x = 2x(x^2 - 4)\).
Now factor the remaining piece. \(x^2 - 4\) is a difference of squares.
\(2x(x^2 - 4) = 2x(x - 2)(x + 2)\). Done.

Missing the GCF first is the number-one factoring mistake we see in class. Always pull it out before trying anything else.

3. Factors, zeros, and \(x\)-intercepts

Once a polynomial is factored, finding its zeros is a one-line move. If \(f(x) = (x - 2)(x - 3)\), the zeros are the values of \(x\) that make \(f(x) = 0\). Since a product is zero when any factor is zero:

\((x - 2)(x - 3) = 0 \quad\Longleftrightarrow\quad x = 2 \text{ or } x = 3\)

Those same values are the \(x\)-intercepts of the graph — the points where the parabola \(y = x^2 - 5x + 6\) crosses the \(x\)-axis: \((2, 0)\) and \((3, 0)\). This equivalence is the reason factoring matters in Algebra II: factors, zeros, roots, solutions, and \(x\)-intercepts are five different names for the same information.

4. The Remainder Theorem

Remainder Theorem

When a polynomial \(f(x)\) is divided by \((x - a)\), the remainder is \(f(a)\).

\(f(x) \div (x - a) \;\longrightarrow\; \text{remainder} = f(a)\)

This one theorem replaces polynomial long division for finding remainders. Instead of setting up the whole long-division tableau, we just plug \(a\) into the polynomial.

Worked example. Find the remainder when \(f(x) = x^3 + 2x^2 - 5x + 3\) is divided by \(x - 2\).

Identify \(a\). Since \(x - 2 = x - a\), we have \(a = 2\).
Evaluate \(f(2) = 2^3 + 2(2^2) - 5(2) + 3 = 8 + 8 - 10 + 3 = 9\).
The remainder is \(\mathbf{9}\) — no long division needed.

5. The Factor Theorem

Factor Theorem

\((x - a)\) is a factor of \(f(x)\) if and only if \(f(a) = 0\).

\((x - a) \text{ is a factor of } f(x) \;\Longleftrightarrow\; f(a) = 0\)

The Factor Theorem is the Remainder Theorem when the remainder happens to be zero. If plugging \(a\) into the polynomial gives zero, then \((x - a)\) divides evenly — so it is a factor. And vice versa.

Worked example. Is \((x - 2)\) a factor of \(f(x) = x^3 - 4x^2 + x + 6\)?

Calculate \(f(2) = 2^3 - 4(2^2) + 2 + 6 = 8 - 16 + 2 + 6 = 0\).
Since \(f(2) = 0\), yes — \((x - 2)\) is a factor of \(f(x)\).

6. The sign trap in \(x + a\) factors

Rule to memorize — always use the OPPOSITE sign

The Factor Theorem is stated for factors of the form \((x - a)\). When the factor is \((x + a)\), rewrite it as \((x - (-a))\) and test \(f(-a)\).

Factor	Test value
\(x - 3\)	\(f(3)\)
\(x + 3\)	\(f(-3)\)
\(x + 4\)	\(f(-4)\)
\(x - 5\)	\(f(5)\)

Every year, half the mistakes on a Factor-Theorem test come from testing \(f(3)\) for the factor \((x + 3)\). Say the rule out loud when you set up each problem: “Factor \(x + 3\), so I test \(f(-3)\).” Always use the opposite of the number you see in the factor.

7. Completely factoring a cubic (Factor Theorem + synthetic division)

Here is where the Factor Theorem earns its keep. A cubic like \(f(x) = x^3 - 6x^2 + 11x - 6\) does not factor with any of our five standard tools. But we can chase down a factor one substitution at a time:

Guess a small \(a\) — usually a factor of the constant term. Here the constant is \(-6\), so try \(\pm 1, \pm 2, \pm 3, \pm 6\).
Try \(x = 1\): \(f(1) = 1 - 6 + 11 - 6 = 0\). By the Factor Theorem, \((x - 1)\) is a factor.
Divide by \((x - 1)\) using synthetic division. The quotient is \(x^2 - 5x + 6\), remainder \(0\).
Factor the quadratic: \(x^2 - 5x + 6 = (x - 2)(x - 3)\).
Write the complete factorization: \(f(x) = (x - 1)(x - 2)(x - 3)\).
Read off the zeros: \(x = 1, \; x = 2, \; x = 3\).

Where the guesses come from. The Rational Root Theorem (Class 12 next week) tells us the only rational zeros of a polynomial with integer coefficients are ratios of factors of the constant term over factors of the leading coefficient. That’s why we start with the divisors of \(-6\) — those are the only candidates worth testing.

8. The Algebra II big picture

Class 11 is really about one chain of ideas that runs through all of Algebra II and every math course after it:

Polynomial → Factor → Zero → \(x\)-intercept

The polynomial is an algebraic expression in \(x\).
A factor writes the polynomial as a product of simpler pieces.
A zero is a value of \(x\) that makes the polynomial equal zero — you get zeros by setting each factor to zero.
An \(x\)-intercept is where the graph crosses the \(x\)-axis — the point \((a, 0)\) for each zero \(a\).

Once a student can move fluently across those four representations, the second half of Algebra II opens up: rational functions, polynomial inequalities, end behavior, and the Fundamental Theorem of Algebra all sit on top of this chain.

Class 11 print worksheet — 25 questions with answer key

A clean 7-page PDF with all 25 practice questions (5 warm-up factoring, 7 mixed factoring + zeros, 5 Remainder Theorem, 5 Factor-Theorem sign-trap, 3 complete-cubic problems) plus a 2-column answer key. Free for classroom + home use.

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9. Practice — 25 questions with hidden answers

Try each one on paper first. Click Answer only after your student has written a full attempt. Questions increase in difficulty: 1–5 warm-up factoring, 6–12 the five techniques, 13–17 Remainder Theorem, 18–22 Factor Theorem (including the sign trap), 23–25 completely factor a cubic.

Print-friendly worksheet — 25 questions, 5 per page

Rather do these on paper? Download all 25 questions on 5 sheets (5 per page) with lines to write the answers, plus an answer key on the last page.

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Warm-up — recognizing the pattern (1–5)

1GCF

Factor completely: \(12x^3 + 18x^2\).

\(6x^2(2x + 3)\).

Greatest common factor of \(12x^3\) and \(18x^2\): number GCF is \(6\); variable GCF is \(x^2\). Pull out \(6x^2\).

2Trinomial

Factor: \(x^2 + 7x + 12\).

\((x + 3)(x + 4)\).

Find two numbers that multiply to \(12\) and add to \(7\). Those are \(3\) and \(4\).

3Diff of squares

Factor: \(x^2 - 25\).

\((x - 5)(x + 5)\).

Recognize \(a^2 - b^2\) with \(a = x\) and \(b = 5\).

4Perfect square

Factor: \(x^2 - 10x + 25\).

\((x - 5)^2\).

Both \(x^2\) and \(25\) are perfect squares, and \(-10x = -2 \cdot x \cdot 5\). That’s the perfect-square-trinomial pattern.

5Trinomial

Factor: \(x^2 - 3x - 10\).

\((x - 5)(x + 2)\).

Two numbers that multiply to \(-10\) and add to \(-3\): \(-5\) and \(2\).

The five techniques — more practice (6–12)

6GCF + trinomial

Factor completely: \(2x^2 + 10x + 12\).

\(2(x + 2)(x + 3)\).

GCF first: \(2(x^2 + 5x + 6)\). Then factor the trinomial inside.

7GCF + diff of squares

Factor completely: \(3x^3 - 12x\).

\(3x(x - 2)(x + 2)\).

GCF is \(3x\). What’s left is \(x^2 - 4\), a difference of squares.

8Grouping

Factor by grouping: \(x^3 + 2x^2 + 3x + 6\).

\((x + 2)(x^2 + 3)\).

Group: \((x^3 + 2x^2) + (3x + 6) = x^2(x + 2) + 3(x + 2)\). Common binomial \((x + 2)\) factors out.

9Trinomial with a

Factor: \(2x^2 + 7x + 3\).

\((2x + 1)(x + 3)\).

Multiply \(a \cdot c = 2 \cdot 3 = 6\). Two numbers multiplying to \(6\) and adding to \(7\): \(1\) and \(6\). Split: \(2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (2x+1)(x+3)\).

10Perfect square

Factor: \(9x^2 + 12x + 4\).

\((3x + 2)^2\).

\(\sqrt{9x^2} = 3x\) and \(\sqrt{4} = 2\). Middle term \(2 \cdot 3x \cdot 2 = 12x\) — matches. Perfect square trinomial.

11Diff of squares

Factor: \(49 - x^2\).

\((7 - x)(7 + x)\).

Same pattern, just written with the constant first. \(a = 7\), \(b = x\).

12Zeros

Find the zeros of \(f(x) = x^2 + 4x - 5\).

\(x = -5\) and \(x = 1\).

Factor: two numbers multiplying to \(-5\) and adding to \(4\) are \(5\) and \(-1\). So \(f(x) = (x + 5)(x - 1)\). Set each factor to zero.

Remainder Theorem (13–17)

13Remainder

Use the Remainder Theorem to find the remainder when \(f(x) = x^3 - 4x^2 + x + 6\) is divided by \(x - 3\).

\(f(3) = 0\). Remainder is \(0\).

\(f(3) = 27 - 36 + 3 + 6 = 0\). So \((x - 3)\) divides evenly. (Bonus — because the remainder is \(0\), \((x-3)\) is a factor by the Factor Theorem.)

14Remainder

Find the remainder when \(f(x) = 2x^3 + 3x^2 - x + 5\) is divided by \(x - 1\).

\(9\).

\(f(1) = 2 + 3 - 1 + 5 = 9\).

15Remainder

Find the remainder when \(f(x) = x^4 - 3x^2 + 2\) is divided by \(x + 2\).

\(6\).

For factor \(x + 2\), test \(f(-2)\). \(f(-2) = 16 - 12 + 2 = 6\).

16Remainder

Find the remainder when \(f(x) = x^3 - 2x + 4\) is divided by \(x + 1\).

\(5\).

Test \(f(-1) = -1 + 2 + 4 = 5\).

17Remainder + interpret

Given \(f(x) = x^3 + kx^2 - 4x + 12\), find the value of \(k\) so that dividing by \(x - 2\) leaves a remainder of \(0\).

\(k = -3\).

By the Remainder Theorem, remainder \(= f(2)\). Compute: \(f(2) = 2^3 + k(2^2) - 4(2) + 12 = 8 + 4k - 8 + 12 = 4k + 12\). Set the remainder to \(0\): \(4k + 12 = 0 \Rightarrow k = -3\). Check: with \(k = -3\), \(f(2) = 8 - 12 - 8 + 12 = 0\). ✓

Factor Theorem & the sign trap (18–22)

18Factor test

Is \((x - 1)\) a factor of \(f(x) = x^3 - x^2 - 4x + 4\)?

Yes.

\(f(1) = 1 - 1 - 4 + 4 = 0\). Since \(f(1) = 0\), \((x - 1)\) is a factor.

19Sign trap

Is \((x + 3)\) a factor of \(f(x) = x^3 + 3x^2 - 4x - 12\)?

Yes.

For factor \(x + 3\), test \(f(-3)\) — opposite sign. \(f(-3) = -27 + 27 + 12 - 12 = 0\). So \((x + 3)\) is a factor.

20Sign trap

Is \((x + 1)\) a factor of \(f(x) = x^3 + 3x^2 - 4x - 12\)?

No.

Test \(f(-1) = -1 + 3 + 4 - 12 = -6 \ne 0\). Since \(f(-1) \ne 0\), \((x + 1)\) is not a factor.

21Sign trap

What value should you substitute into \(f(x)\) to check whether \((x + 7)\) is a factor?

\(f(-7)\).

Factor \(x + 7\) is the same as \(x - (-7)\), so \(a = -7\).

22Find a factor

Given \(f(x) = x^3 - 7x + 6\), test small integer values and find one linear factor.

\((x - 1)\) is a factor (also \((x - 2)\) and \((x + 3)\)).

Divisors of the constant \(6\): \(\pm 1, \pm 2, \pm 3, \pm 6\). Test \(f(1) = 1 - 7 + 6 = 0\) ✓. Bonus: \(f(2) = 8 - 14 + 6 = 0\) ✓ and \(f(-3) = -27 + 21 + 6 = 0\) ✓ — you have all three factors of the cubic.

Completely factor — putting it all together (23–25)

23Cubic

Completely factor \(f(x) = x^3 - 6x^2 + 11x - 6\) and find its zeros.

\(f(x) = (x - 1)(x - 2)(x - 3)\). Zeros: \(x = 1, 2, 3\).

Try \(x = 1\): \(f(1) = 1 - 6 + 11 - 6 = 0\) — so \((x - 1)\) is a factor. Synthetic division by \(1\): coefficients \(1, -6, 11, -6\) → quotient \(x^2 - 5x + 6 = (x - 2)(x - 3)\).

24Cubic

Completely factor \(f(x) = x^3 + 2x^2 - 5x - 6\) and find its zeros.

\(f(x) = (x - 2)(x + 1)(x + 3)\). Zeros: \(x = 2, -1, -3\).

Test candidates. \(f(2) = 8 + 8 - 10 - 6 = 0\) ✓. Divide by \((x - 2)\) synthetically: quotient \(x^2 + 4x + 3 = (x + 1)(x + 3)\).

25Challenge

Completely factor \(f(x) = 2x^3 - x^2 - 8x + 4\) and find its zeros.

\(f(x) = (2x - 1)(x - 2)(x + 2)\). Zeros: \(x = \tfrac{1}{2}, \; 2, \; -2\).

By the Rational Root Theorem (Class 12 preview), candidates are \(\pm 1, \pm 2, \pm 4, \pm \tfrac{1}{2}\). Test \(f(2) = 16 - 4 - 16 + 4 = 0\) ✓. Divide by \((x - 2)\) synthetically: quotient \(2x^2 + 3x - 2 = (2x - 1)(x + 2)\).

10. FAQ for parents

What should my child already know before this class?

Fluency with the distributive property, multiplying binomials, and factoring simple trinomials like \(x^2 + 5x + 6\). If your student is comfortable with our Pre-Algebra Class 12 on expressions and the Algebra 1 quadratics unit, they are ready.

Where does this show up on tests?

The Factor Theorem and Remainder Theorem appear directly on the Algebra II Regents, the SAT (Advanced Math domain), and every precalculus placement test. In our own SAT sessions they show up disguised as “which of the following is a factor of the polynomial\(\ldots\)” multiple-choice questions.

How do I help without knowing the material?

Ask your student to say out loud what test value the factor tells them to plug in. If they can articulate “for factor \(x + 3\), I test \(f(-3)\),” they have the sign trap under control.

11. What’s next

Young Fermats Algebra II is a rolling honors course — join at any class. Class 12 next week is Zeros of Polynomials — the Rational Root Theorem and end behavior, which extends everything above to fourth- and fifth-degree polynomials. Ask us at your free 30-minute evaluation whether your student is ready to join Algebra II now or start with Algebra 1 Ignite first.

Related classes on the site:

Come see us on the Upper West Side. SOMATH is at 226 W 79th Street. Call (646) 668-6151 or book a free 30-minute evaluation to see whether Young Fermats Algebra II is the right fit for your student this term.