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➕ Algebra 1 · Factoring

Factoring Trinomials

Factor trinomials like x^2 + bx + c and ax^2 + bx + c with sum-and-product and the ac method, check by expanding, and spot prime and perfect square cases.

Grades 9, 10⏱ 16 min✔ AI-checkedCCSS A-SSE.B.3aCCSS A-SSE.A.2
About standards and checking

Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Two blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-08.

  • A-SSE.B.3a: Factor a quadratic expression to reveal the zeros of the function it defines.
  • A-SSE.A.2: Use the structure of an expression to identify ways to rewrite it.

By the end you can…

  • Factor x2 + bx + c by finding two integers whose product is c and whose sum is b, using the signs of b and c to guide the search.
  • Factor ax2 + bx + c with a ≠ 1 by the ac method (split the middle term, then group), always taking out the greatest common factor first.
  • Check every factorization by expanding, and recognize trinomials that cannot be factored using integers (prime).
  • Recognize perfect square trinomials, and explain how a factored form reveals the zeros.

Lesson (school style)

A trinomial is a polynomial with three terms, like x2 + 11x + 28. Multiplying two binomials often produces one, and factoring a trinomial runs that multiplication backward: you start with the trinomial and find the two binomials. The key is a pattern. In (x + p)(x + q), the numbers p and q multiply to the constant term and add to the coefficient of x. Take out any GCF first, and always check by expanding.

Factoring is un-multiplying

Expand (x + 4)(x + 7): x2 + 7x + 4x + 28 = x2 + 11x + 28. Look at the numbers 4 and 7. They multiply to 28, the constant term, and they add to 11, the coefficient of x.

That is no accident. For any numbers p and q,

(x + p)(x + q) = x2 + qx + px + pq = x2 + (p + q)x + pq.

The middle coefficient is p + q and the constant term is pq. So to factor x2 + bx + c (a trinomial in standard form, descending powers, with a coefficient of 1 on x2), look for two integers p and q with pq = c and p + q = b. Then x2 + bx + c = (x + p)(x + q).

The diagram shows why for positive numbers: an (x + 4) by (x + 7) rectangle splits into an x2 piece, a 7x piece, a 4x piece and a 28 piece.

Step by step when the leading coefficient is 1

  1. Write the trinomial in standard form and take out any common factor first. This method needs a coefficient of 1 on x2.
  2. List pairs of integers whose product is c.
  3. Find the pair whose sum is b.
  4. Write (x + p)(x + q).
  5. Check by expanding. You must get the original trinomial exactly.

The check matters. A pair with the right product but the wrong signs gives the right constant term and the wrong middle term, and only expanding shows it.

Let the signs guide you

Signs tell you where to look. These rules are for x2 + bx + c when b and c are not 0.

  • c is positive. Then p and q have the same sign, because a positive product needs matching signs. The sign of b tells you which: both positive if b is positive, both negative if b is negative. Example: x2 + 8x + 15 = (x + 3)(x + 5), and x2 − 7x + 10 = (x − 2)(x − 5).
  • c is negative. Then p and q have opposite signs, because a negative product needs different signs. A sum takes the sign of the number that is larger in size, so if b is positive the larger one is positive, and if b is negative the larger one is negative. Example: x2 + 2x − 15 = (x + 5)(x − 3), and x2 − 3x − 18 = (x − 6)(x + 3).

Use these rules to skip pairs that cannot work.

Always take out the GCF first

If all three terms share a common factor, pull it out first, as in the last lesson. That often leaves a trinomial with a coefficient of 1 on x2:

3x2 + 18x + 24 = 3(x2 + 6x + 8) = 3(x + 2)(x + 4).

The GCF stays in the answer as a factor. Dropping it gives (x + 2)(x + 4), which expands to x2 + 6x + 8, only a third of the original.

If the leading coefficient is negative, also factor out the negative of the GCF, so that the x2 term inside is positive. For −x2 + 5x + 14 the GCF is 1, and the negative of it is −1: −x2 + 5x + 14 = −(x2 − 5x − 14) = −(x − 7)(x + 2). Notice that every sign inside flips when you factor out a negative.

An expression is completely factored when the GCF is out and no remaining factor can be factored further using integers. (A number out front, such as 3 or 4, counts as one factor.) Check each factor. For example, (4x + 8)(x + 5) expands to 4x2 + 28x + 40, but 4x + 8 still has the common factor 4. The complete answer is 4(x + 2)(x + 5).

When the leading coefficient is not 1: the ac method

For ax2 + bx + c with a ≠ 1, you cannot just look for numbers with product c. Here is why. Multiply two general binomials: (rx + s)(tx + u) = rtx2 + (ru + st)x + su. The leading coefficient is a = rt, the constant is c = su, and the middle coefficient is b = ru + st. The two cross products ru and st add to b, and they multiply to (ru)(st) = (rt)(su) = ac. So look for two integers with product ac and sum b. This is the ac method.

Remember it as SPLICE:

  • Strip out any GCF first.
  • Product: compute a · c.
  • Look for two integers whose product is ac and whose sum is b.
  • Insert them: rewrite bx as two x-terms that use those integers.
  • Common factors: factor the GCF out of the first two terms and out of the last two terms. The same binomial should appear in both. Pull it out.
  • Expand to check.

Tips. Use the sign rules from “Let the signs guide you” with ac in place of c. If the two groups do not show the same binomial, recheck your signs. When the second pair begins with a negative term, factor out a negative GCF so the binomials match. Writing the two split terms in the other order does not change the final answer, as long as you take the full GCF out of each pair. And when a = 1 this is the same method as before, because ac = c.

Other methods exist

Many teachers also use trial and check: list the ways to get the first term and the last term, try them in binomials, and test which one gives the right middle term. Others use the box (area) method or slide and divide. All of them find the same factorization. This lesson uses the ac method because it is systematic and does not depend on a lucky guess. Use whichever method you can do accurately, and always check by expanding.

Trinomials that do not factor: prime

Sometimes no pair of integers works. Try x2 + 3x + 5. The integer pairs with product 5 are 1 · 5 (sum 6) and (−1)(−5) (sum −6). Neither sum is 3, so no integer pair fits. The trinomial is prime (cannot be factored using integers). Say so and stop. Do not force an answer. For ax2 + bx + c with any GCF already taken out, the same idea applies: if no pair of integers has product ac and sum b, the trinomial is prime.

List every integer pair for the product before you decide, including negative pairs when the signs allow them.

Prime only means “cannot be factored using integers.” It does not mean the expression is useless. Later you will meet other tools, such as the quadratic formula, for finding where a trinomial equals 0 (or showing that it never does), whether or not it factors using integers.

A quick test is coming in the discriminant lesson: when a, b and c are integers and any GCF has been taken out, the trinomial factors into two binomials with integer coefficients exactly when b2 − 4ac is a perfect square (0, 1, 4, 9, 16, and so on). For x2 + 3x + 5 it is 9 − 20 = −11, which is not a perfect square, so the trinomial is prime.

Missing a term?

  • No middle term (b = 0). Then you have a binomial, not a trinomial, but the same search works. You need p + q = 0, so q = −p. For x2 − 49 the product is −49 and the pair is 7 and −7, so x2 − 49 = (x − 7)(x + 7). This pattern has its own lesson, difference of squares. But x2 + 49 is prime: a positive product needs two numbers with the same sign, and two numbers with the same sign cannot add to 0.
  • No constant term (c = 0). Then you do not have a trinomial. x2 + 5x factors by taking out the GCF: x(x + 5).

Perfect square trinomials

Sometimes the two numbers are equal, so the factors repeat: x2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)2 and x2 − 10x + 25 = (x − 5)2. These are perfect square trinomials. For any numbers u and v: u2 + 2uv + v2 = (u + v)2 and u2 − 2uv + v2 = (u − v)2.

To recognize one, check two things: the first and last terms are perfect squares, and the middle term is twice the product of their square roots (with a plus or minus sign). For 4x2 + 12x + 9: 4x2 = (2x)2, 9 = 32, and 2 · 2x · 3 = 12x. So it equals (2x + 3)2.

Both end terms being squares is not enough. The trinomial x2 + 12x + 25 has x2 and 25 = 52, but 2 · x · 5 = 10x, not 12x. It is not a perfect square trinomial (in fact it is prime).

Sum-and-product and the ac method still work on these, because the two numbers just happen to be equal. Spotting the pattern is a shortcut. The Special Products of Binomials lesson shows the pattern from the multiplying side.

What factoring reveals: the zeros

Once a trinomial is factored, you can read off when it equals 0. A product of numbers is 0 only when at least one factor is 0. So (x − 3)(x + 7) = 0 exactly when x = 3 or x = −7. These x-values are the zeros of the function f(x) = (x − 3)(x + 7).

This is what the standard means by factoring a quadratic “to reveal the zeros.” Notice the sign flip: the factor (x + 7) gives the zero x = −7. Solving whole equations this way has its own lesson, Solving Quadratics by Factoring.

Key words

Trinomial
A polynomial with three terms, such as x2 + 11x + 28. A binomial has two terms and a monomial has one.
Standard form
A polynomial written with its terms in descending powers, such as 2x2 − 3x − 20.
Leading coefficient
The coefficient of the term with the highest power, when written in standard form. In 2x2 − 3x − 20 it is 2.
Degree
The highest exponent of the variable in a polynomial. A trinomial like x2 + 11x + 28 has degree 2, so it is called quadratic.
Factoring
Rewriting an expression as a product. For trinomials, it is multiplying two binomials run backward.
Completely factored
Written as a product in which the GCF is out and no factor can be factored further using integers.
Prime (cannot be factored using integers)
A polynomial that cannot be written as a product of simpler polynomials with integer coefficients, other than by taking out 1 or −1, such as x2 + 3x + 5. A trinomial with a GCF greater than 1, such as 7x2 + 14x + 21, is not prime: take out the GCF and test what is left.
ac method
A way to factor ax2 + bx + c: find two integers with product ac and sum b, split the middle term with them, then factor by grouping.
Perfect square trinomial
A trinomial that is the square of a binomial, such as x2 + 6x + 9 = (x + 3)2. The pattern is u2 ± 2uv + v2 = (u ± v)2.
Zero of a function
An input value that makes the output 0. The zeros of f(x) = (x − 3)(x + 7) are 3 and −7.

Worked examples

Example 1. Factor x2 + 8x + 15.

  1. The coefficient of x2 is 1 and the terms share no common factor, so look for two integers with product c = 15 and sum b = 8.
  2. Since 15 is positive, the two numbers have the same sign. Since 8 is positive, both are positive.
  3. List the pairs with product 15: 1 · 15 has sum 16, and 3 · 5 has sum 8.
  4. The pair 3 and 5 works, so x2 + 8x + 15 = (x + 3)(x + 5).
  5. Check by expanding: (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15.

Answer: Answer: (x + 3)(x + 5)

Example 2. Factor x2 − 9x + 20.

  1. c = 20 is positive, so the two numbers have the same sign. b = −9 is negative, so both are negative.
  2. Negative pairs with product 20: (−1)(−20) has sum −21, (−2)(−10) has sum −12, and (−4)(−5) has sum −9.
  3. The pair −4 and −5 works, so x2 − 9x + 20 = (x − 4)(x − 5).
  4. Check: (x − 4)(x − 5) = x2 − 5x − 4x + 20 = x2 − 9x + 20. The positive pair 4 and 5 also has product 20, but (x + 4)(x + 5) would give +9x, not −9x.

Answer: Answer: (x − 4)(x − 5)

Example 3. Factor x2 + 2x − 15.

  1. c = −15 is negative, so the numbers have opposite signs. b = 2 is positive, so the larger one (in size) is positive.
  2. Pairs with product −15 and the positive number larger: (−1)(15) has sum 14, and (−3)(5) has sum 2.
  3. The pair −3 and 5 works, so x2 + 2x − 15 = (x + 5)(x − 3).
  4. Check: (x + 5)(x − 3) = x2 − 3x + 5x − 15 = x2 + 2x − 15.

Answer: Answer: (x + 5)(x − 3)

Example 4. Factor x2 − 3x − 18.

  1. c = −18 is negative, so the numbers have opposite signs. b = −3 is negative, so the larger one (in size) is negative.
  2. Pairs with product −18 and the negative number larger: (−18)(1) has sum −17, (−9)(2) has sum −7, and (−6)(3) has sum −3.
  3. The pair −6 and 3 works, so x2 − 3x − 18 = (x − 6)(x + 3).
  4. Check: (x − 6)(x + 3) = x2 + 3x − 6x − 18 = x2 − 3x − 18.

Answer: Answer: (x − 6)(x + 3)

Example 5. Factor 3x2 + 18x + 24 completely.

  1. Take out the GCF first. The GCF of 3, 18 and 24 is 3: 3x2 + 18x + 24 = 3(x2 + 6x + 8).
  2. Inside, c = 8 and b = 6 are both positive, so look for two positive integers with product 8 and sum 6. The pairs are 1 · 8 (sum 9) and 2 · 4 (sum 6).
  3. So x2 + 6x + 8 = (x + 2)(x + 4). Keep the 3: 3(x + 2)(x + 4).
  4. Check: (x + 2)(x + 4) = x2 + 6x + 8, and 3(x2 + 6x + 8) = 3x2 + 18x + 24. Leaving out the 3 would give only x2 + 6x + 8.
  5. The factors x + 2 and x + 4 are linear binomials, so neither can be factored further. The 3 out front counts as one factor.

Answer: Answer: 3(x + 2)(x + 4)

Example 6. Factor −x2 + 5x + 14 completely.

  1. The leading coefficient is negative, so factor out −1. Every sign inside flips: −x2 + 5x + 14 = −(x2 − 5x − 14).
  2. Inside, c = −14 is negative, so the numbers have opposite signs. b = −5 is negative, so the larger one is negative. Pairs: (−14)(1) has sum −13, and (−7)(2) has sum −5.
  3. So x2 − 5x − 14 = (x − 7)(x + 2), and −x2 + 5x + 14 = −(x − 7)(x + 2).
  4. Check: (x − 7)(x + 2) = x2 + 2x − 7x − 14 = x2 − 5x − 14. Multiplying by −1 gives −x2 + 5x + 14.

Answer: Answer: −(x − 7)(x + 2)

Example 7. Factor 6x2 + 11x + 3 with the ac method.

  1. S: the terms 6, 11 and 3 share only the factor 1, so there is no GCF to take out.
  2. P: a · c = 6 × 3 = 18.
  3. L: find two integers with product 18 and sum 11. Both are positive. The pairs are 1 · 18 (sum 19), 2 · 9 (sum 11) and 3 · 6 (sum 9). The pair is 2 and 9.
  4. I: split the middle term: 6x2 + 2x + 9x + 3.
  5. C: the first two terms have GCF 2x, so 6x2 + 2x = 2x(3x + 1). The last two have GCF 3, so 9x + 3 = 3(3x + 1). The same binomial (3x + 1) appears, so 2x(3x + 1) + 3(3x + 1) = (3x + 1)(2x + 3).
  6. E: (3x + 1)(2x + 3) = 6x2 + 9x + 2x + 3 = 6x2 + 11x + 3.
  7. Splitting in the other order gives the same factors: 6x2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1).

Answer: Answer: (3x + 1)(2x + 3)

Example 8. Factor 5x2 − 12x − 9.

  1. S: 5, 12 and 9 share only the factor 1. P: a · c = 5 × (−9) = −45.
  2. L: the product −45 is negative, so the numbers have opposite signs. The sum −12 is negative, so the larger one is negative. Pairs: (−45)(1) has sum −44, (−15)(3) has sum −12, and (−9)(5) has sum −4. The pair is −15 and 3.
  3. I: split −12x into −15x + 3x: 5x2 − 15x + 3x − 9.
  4. C: 5x2 − 15x = 5x(x − 3) and 3x − 9 = 3(x − 3). The common binomial is (x − 3), so the result is (x − 3)(5x + 3).
  5. E: (x − 3)(5x + 3) = 5x2 + 3x − 15x − 9 = 5x2 − 12x − 9.

Answer: Answer: (x − 3)(5x + 3)

Example 9. Factor 12x2 − 26x + 10 completely.

  1. S: the GCF of 12, 26 and 10 is 2, so 12x2 − 26x + 10 = 2(6x2 − 13x + 5).
  2. P: for the trinomial inside, a · c = 6 × 5 = 30.
  3. L: c = 5 is positive and b = −13 is negative, so both numbers are negative. Pairs with product 30: (−1)(−30) has sum −31, (−2)(−15) has sum −17, (−3)(−10) has sum −13, and (−5)(−6) has sum −11. The pair is −3 and −10.
  4. I: 6x2 − 3x − 10x + 5.
  5. C: 6x2 − 3x = 3x(2x − 1). For the last two terms, factor out a negative GCF so the binomial matches: −10x + 5 = −5(2x − 1). Then 3x(2x − 1) − 5(2x − 1) = (2x − 1)(3x − 5).
  6. Keep the 2 from step S: 2(2x − 1)(3x − 5).
  7. E: (2x − 1)(3x − 5) = 6x2 − 10x − 3x + 5 = 6x2 − 13x + 5, and 2(6x2 − 13x + 5) = 12x2 − 26x + 10.

Answer: Answer: 2(2x − 1)(3x − 5)

Example 10. Decide whether each trinomial can be factored using integers: (a) x2 + 6x + 4 and (b) 3x2 + 5x + 4.

  1. (a) b = 6 and c = 4 are both positive, so any pair must be positive. The positive pairs with product 4 are 1 · 4 (sum 5) and 2 · 2 (sum 4). Negative pairs have negative sums. No pair adds to 6.
  2. (a) So x2 + 6x + 4 is prime (cannot be factored using integers).
  3. (b) There is no GCF. a · c = 3 × 4 = 12 and b = 5 is positive, so look for positive pairs with product 12: 1 · 12 (sum 13), 2 · 6 (sum 8) and 3 · 4 (sum 7). None adds to 5.
  4. (b) So 3x2 + 5x + 4 is also prime.

Answer: Answer: Both are prime: neither can be factored using integers.

Example 11. Factor 4x2 + 20x + 25. Then decide whether x2 + 10x + 16 is also a perfect square trinomial.

  1. 4x2 = (2x)2 and 25 = 52. Twice the product of 2x and 5 is 2 · 2x · 5 = 20x, which is exactly the middle term. So it is a perfect square trinomial: (2x + 5)2.
  2. Check: (2x + 5)(2x + 5) = 4x2 + 10x + 10x + 25 = 4x2 + 20x + 25.
  3. Now x2 + 10x + 16. The end terms are squares (x2 and 42), but 2 · x · 4 = 8x, not 10x. It is not a perfect square trinomial.
  4. Factor it with sum and product instead: product 16 and sum 10 gives 2 and 8, so (x + 2)(x + 8). Check: x2 + 8x + 2x + 16 = x2 + 10x + 16.

Answer: Answer: 4x2 + 20x + 25 = (2x + 5)2. The trinomial x2 + 10x + 16 is not a perfect square trinomial: it equals (x + 2)(x + 8).

Example 12. Factor x2 + 4x − 21 and find the x-values that make it equal 0.

  1. c = −21 is negative, so the numbers have opposite signs. b = 4 is positive, so the larger one is positive. Pairs: (−1)(21) has sum 20, and (−3)(7) has sum 4.
  2. So x2 + 4x − 21 = (x − 3)(x + 7). Check: x2 + 7x − 3x − 21 = x2 + 4x − 21.
  3. A product is 0 only if one of its factors is 0. So x − 3 = 0 gives x = 3, and x + 7 = 0 gives x = −7.
  4. Check both: 32 + 4(3) − 21 = 9 + 12 − 21 = 0 and (−7)2 + 4(−7) − 21 = 49 − 28 − 21 = 0.

Answer: Answer: (x − 3)(x + 7). The expression equals 0 when x = 3 or x = −7.

In plain English

Multiplying two small expressions gives a longer one. For example, (x + 4)(x + 7) = x2 + 11x + 28. Factoring a trinomial goes backward. You start with the three-part expression and find the two small ones.

There is a pattern to spot. Look at the 4 and the 7. They multiply to 28, the last number. They add to 11, the number in front of x. So to factor x2 + 11x + 28, ask: which two numbers multiply to 28 and add to 11?

If the number in front of x2 is not 1, there is one extra step. Multiply that number by the last number. Find two numbers that make that product and still add to the middle number. Then split the middle part into two parts and group.

Take out any common factor first. Always check by multiplying your answer back out. Some trinomials cannot be factored using integers (whole numbers and their negatives). Those are called prime.

See it

A rectangle (x + 4) by (x + 7) shows the pattern x7 x4 x2x · x 7xx · 7 4x4 · x 284 · 7 They add to b 4 + 7 = 11 7x + 4x = 11x, so b = 11 They multiply to c 4 × 7 = 28 the corner piece, so c = 28 (x + 4)(x + 7) x2 + 11x + 28 expandfactor Drawn for positive x and not to scale. The algebra works for every number.
In an (x + 4) by (x + 7) rectangle, the two middle pieces 7x and 4x add to 11x, and the corner piece is 4 · 7 = 28. So x2 + 11x + 28 = (x + 4)(x + 7).
The ac method (SPLICE) on 6x2 + 11x + 3 S Strip the GCF common factor of all three terms? none: 6, 11 and 3 share only 1 P Product multiply a · c 6 × 3 = 18 L Look for a pair product 18, sum 11 2 and 9 (2 × 9 = 18, 2 + 9 = 11) I Insert: split 11x into 2x + 9x 6x2 + 2x + 9x + 3 C Common factors of each pair, then the binomial 2x(3x + 1) + 3(3x + 1) = (3x + 1)(2x + 3) E Expand to check multiply the answer out (3x + 1)(2x + 3) = 6x2 + 11x + 3 Same steps for any ax2 + bx + c. If no pair works (GCF already out), the trinomial is prime.
The ac method on 6x2 + 11x + 3: find the pair with product 18 and sum 11 (2 and 9), split the middle term, factor each pair, and pull out the common binomial.

The same idea, the fun way

Pick the one that clicks. Each one explains the same concept from a different angle.

Gaming

The tile-packing puzzle

Picture a puzzle game where you must pack every tile you are given into one perfect rectangle, with no gaps and no overlaps. For x2 + 9x + 14 the pile is one big square (x by x), 9 long strips (x by 1) and 14 little squares (1 by 1). They fit into a rectangle that is (x + 2) by (x + 7). The strips split into 7 along one side and 2 along the other, and the little squares form a 2-by-7 block.

How it connects: The two extra side lengths, 2 and 7, add to the number of strips (2 + 7 = 9) and multiply to the number of little squares (2 · 7 = 14). That is why we look for a pair with sum b and product c. The limit: this picture works only when every term is positive, because tile counts and lengths cannot be negative, and it covers only a = 1. If no rectangle can use all the tiles, the trinomial is prime.

Sports

Net yards

In football, a team can gain yards on one play and lose yards on another. Say one play gains 8 yards and another loses 5. Together they give a net gain of 3 yards, because the bigger play, the gain, decides whether the total is a gain or a loss.

How it connects: To factor x2 + 3x − 40 you need two numbers with product −40 and sum +3. A negative product means opposite signs, and a positive sum means the larger number (in size) is positive. The pair is 8 and −5, so x2 + 3x − 40 = (x + 8)(x − 5). The limit: the yards story explains only why the sign of the sum follows the larger number. Yards have no product, so the story cannot find the numbers for you. You still list pairs with product −40.

Food

The double batch

A bakery writes a recipe for a double batch, so every amount is a multiple of 2. The head baker rewrites it as “2 times one batch.” Say the double batch is 2x2 + 14x + 24. One batch is x2 + 7x + 12, and that smaller recipe is the one worth factoring: (x + 3)(x + 4).

How it connects: 2x2 + 14x + 24 = 2(x2 + 7x + 12) = 2(x + 3)(x + 4). The 2 is the GCF, and it stays in the final answer, like the number of batches. If you drop it, (x + 3)(x + 4) expands to x2 + 7x + 12, only half of the order. The limit: you can only pull out a batch count that divides every term. In 2x2 + 14x + 25 the 2 does not divide 25, so there is no batch count to pull out.

Idiom

A square peg in a round hole

To call something “a square peg in a round hole” means it does not fit where you are trying to put it, no matter how hard you push. Try to factor x2 + 5x + 3. You need two integers with product 3 and sum 5. The only integer pairs with product 3 are 1 · 3 (sum 4) and (−1)(−3) (sum −4). Neither sum is 5.

How it connects: When no integer pair fits, stop pushing: the trinomial is prime (cannot be factored using integers). The limit: prime is about integers only. It does not say the trinomial is useless. The quadratic formula, later in this course, can still find the x-values where x2 + 5x + 3 equals 0. Those values involve square roots.

Proverb

The proof of the pudding is in the eating

This old proverb says you can only judge something by trying it. Suppose you claim x2 + 5x − 24 = (x + 3)(x − 8). To “eat the pudding,” expand your answer.

How it connects: Expanding gives (x + 3)(x − 8) = x2 − 8x + 3x − 24 = x2 − 5x − 24. The middle term is −5x, not +5x, so the claim is wrong: the signs were swapped. The correct factoring is (x + 8)(x − 3). The limit: expanding shows that your answer equals the original. It does not tell you whether you took out the GCF or whether a factor can be factored further, so check those separately.

Everyday joke

Staying single

Why did the trinomial x2 + 3x + 5 stay single? It looked for a pair of integers that multiply to 5 and add to 3, and no pair was a match.

How it connects: The integer pairs with product 5 are 1 · 5 (sum 6) and (−1)(−5) (sum −6). Neither sum is 3, so x2 + 3x + 5 is prime (cannot be factored using integers). The limit: prime only says that integer binomials will not work. It does not mean anything is wrong with the trinomial.

Memory help

Acronym

SPLICE the trinomial

For ax2 + bx + c: Strip out the GCF first. Product: multiply a · c. Look for two integers with product ac and sum b. Insert them to split the middle term. Common factors: factor each pair, then pull out the common binomial. Expand to check. When a = 1 the same steps work, with ac = c.

Rhyme

Times to c, plus to b

Times to c, plus to b: that is the pair that sets you free. Write the binomials, one and two, then expand to check it’s true. (This is for x2 + bx + c, after taking out any GCF. When a ≠ 1, use ac in place of c.)

Sketch it

The sum-and-product table

Draw a table with two columns. In the first, list the integer pairs whose product is c. In the second, write each pair’s sum. Circle the row whose sum is b. For x2 + 6x − 16 the pairs (−1, 16), (−2, 8) and (−4, 4) have sums 15, 6 and 0, so the pair is −2 and 8 and the answer is (x − 2)(x + 8). Use the sign rules to skip rows that cannot work.

Trap alert

Right product, wrong signs

x2 − 7x + 10 is (x − 2)(x − 5), not (x + 2)(x + 5). Both pairs multiply to 10, but only −2 and −5 add to −7. Expand (x + 2)(x + 5) and you get x2 + 7x + 10: the middle term gives the mistake away.

Trap alert

Do not drop the GCF

6x2 + 12x − 18 = 6(x2 + 2x − 3) = 6(x + 3)(x − 1), not (x + 3)(x − 1). The 6 is a factor of the answer. Expand (x + 3)(x − 1) and you get x2 + 2x − 3, which is only one sixth of the original.

Trap alert

Use ac, not c

To factor 4x2 + 8x + 3, look for two integers with product a · c = 4 · 3 = 12 and sum 8. That pair is 2 and 6. Looking for product 3 and sum 8 finds nothing, and you might wrongly decide the trinomial is prime. It is not: 4x2 + 8x + 3 = (2x + 1)(2x + 3).

Study notes

Key ideas

  • Factoring a trinomial is un-multiplying two binomials. For x2 + bx + c, find integers p and q with pq = c and p + q = b. Then x2 + bx + c = (x + p)(x + q).
  • Signs (for nonzero b and c): when c is positive, p and q have the same sign (the sign of b). When c is negative, they have opposite signs, and the one larger in size has the sign of b.
  • Always take out the GCF first (a negative GCF if the leading coefficient is negative), and keep it in the final answer.
  • When a ≠ 1, use the ac method: find two integers with product ac and sum b, split the middle term, then factor by grouping.
  • If no pair of integers works, the trinomial is prime (cannot be factored using integers). Always check by expanding.
  • Perfect square trinomials, u2 ± 2uv + v2 = (u ± v)2, are the case where the two numbers are equal. A factored form reveals the zeros: a product is 0 only when a factor is 0.

Steps

  1. Write the trinomial in standard form (descending powers). Take out the GCF, using a negative GCF if the leading coefficient is negative.
  2. If the coefficient of x2 is 1: list integer pairs with product c, pick the pair with sum b, and write (x + p)(x + q).
  3. If the coefficient of x2 is not 1: find two integers with product ac and sum b. Split the middle term, factor each pair, and pull out the common binomial.
  4. If no pair works, say the trinomial is prime.
  5. Expand to check. Make sure the GCF is in your final answer and that no factor can be factored further.

Formulas

  • Multiplying two binomials: (x + p)(x + q) = x2 + (p + q)x + pq
  • Factoring $x^2 + bx + c$: x2 + bx + c = (x + p)(x + q) when pq = c and p + q = b
  • ac method: Find integers m and n with mn = ac and m + n = b. Then ax2 + bx + c = ax2 + mx + nx + c, and you factor by grouping.
  • Perfect square trinomials: u2 + 2uv + v2 = (u + v)2 and u2 − 2uv + v2 = (u − v)2

Watch out

  • Right product, wrong signs: expand to catch it. (x + 2)(x + 5) is x2 + 7x + 10, not x2 − 7x + 10.
  • Do not drop the GCF: 6x2 + 12x − 18 = 6(x + 3)(x − 1).
  • When a ≠ 1, look for the product ac, not c.
  • Both end terms being perfect squares does not make a perfect square trinomial. Check that the middle term is twice the product of the square roots.
  • “Prime” only means “cannot be factored using integers.” List every integer pair, including negative ones, before you decide.

Why it works

  • Why do the numbers multiply to c and add to b? Because (x + p)(x + q) = x2 + (p + q)x + pq: expanding adds the two cross terms and multiplies the two constants.
  • Why does the ac method work? In (rx + s)(tx + u) the cross products ru and st add to b and multiply to (rt)(su) = ac. So the two numbers with product ac and sum b are exactly those cross terms. Splitting the middle term into them lets you factor by grouping.
  • Why factor at all? A product is 0 only when one of its factors is 0, so a factored quadratic shows the zeros (the x-values where it equals 0) at a glance.

Common mistakes

✗ Mistakex2 − 7x + 10 = (x + 2)(x + 5) (right product, wrong signs).
✓ InsteadBoth pairs multiply to 10, but the middle term −7x needs −2 and −5: (x − 2)(x − 5). Expanding (x + 2)(x + 5) gives x2 + 7x + 10, which catches the mistake.
✗ Mistake6x2 + 12x − 18 = (x + 3)(x − 1) (dropping the GCF).
✓ InsteadThe GCF 6 is part of the answer: 6(x + 3)(x − 1). Without it, you only have x2 + 2x − 3, one sixth of the original.
✗ MistakeTo factor 4x2 + 8x + 3, look for two numbers with product 3 and sum 8.
✓ InsteadWhen a ≠ 1, use the product ac = 12 and sum 8: the pair is 2 and 6. Then 4x2 + 8x + 3 = (2x + 1)(2x + 3).
✗ Mistakex2 + 12x + 25 is a perfect square trinomial because x2 and 25 are perfect squares.
✓ InsteadThe middle term must be twice the product of the square roots: 2 · x · 5 = 10x, not 12x. So it is not a perfect square trinomial. No integer pair has product 25 and sum 12, so it is prime.
✗ MistakeIf the positive pairs do not work, the trinomial must be prime. For example, x2 − 13x + 36 looks prime.
✓ InsteadCheck negative pairs too. When c is positive and b is negative, both numbers are negative: (−4)(−9) = 36 and −4 + (−9) = −13, so x2 − 13x + 36 = (x − 4)(x − 9).

Teach it back

Explain to a friend how to factor 2x2 − 3x − 20 completely and how to check your answer. Say why you look for two numbers with product −40.

Explaining it out loud shows what you really know. Did you…

  • I checked for a GCF first. There is none: 2, −3 and −20 share only the factor 1.
  • I multiplied a · c = 2 × (−20) = −40 and found two integers with product −40 and sum −3. They are 5 and −8.
  • I explained why: in (rx + s)(tx + u) the two cross products ru and st add to b and multiply to (rt)(su) = ac, so the split must use two numbers with product ac = −40 and sum b = −3.
  • I split the middle term and grouped: 2x2 + 5x − 8x − 20 = x(2x + 5) − 4(2x + 5) = (2x + 5)(x − 4).
  • I expanded to check: (2x + 5)(x − 4) = 2x2 − 8x + 5x − 20 = 2x2 − 3x − 20.

Your 3-week memory plan

  1. TodayRead the lesson, then try the flashcards.
  2. TomorrowRecall it from memory first, then check.
  3. In 3 daysTake the true/false and practice questions.
  4. In 1 weekMix it with older topics and teach it back.
  5. In 3 weeksA last quick review before it fades.

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Factoring Out the Greatest Common FactorMultiplying PolynomialsSpecial Products of BinomialsDifference of SquaresFactoring by GroupingSolving Quadratics by Factoring