4.2 Factoring Quadratic Trinomials

4.2 Factoring Quadratic Trinomials

College Board’s Table 11 factoring skill is methods applied to quadratics, cubics, and polynomials. The workhorse method on AAF is factoring a quadratic trinomial ax^2 + bx + c into two binomials. When a = 1, you hunt an integer pair that multiplies to c and adds to b. When a ≠ 1, you use the ac method (factoring by grouping). Either way, you check by FOIL. If no integer pair exists, the trinomial is prime over the integers — stop factoring and, if the task is to solve, move to the quadratic formula.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

When a = 1: multiply to c, add to b

For x^2 + bx + c, you want integers p and q such that:

• p · q = c
• p + q = b

Then x^2 + bx + c = (x + p)(x + q).

Signs follow the product and the sum:

Worked example: x² + 5x − 24

Find two integers whose product is −24 and whose sum is 5.

Pairs that multiply to −24 include (1, −24), (2, −12), (3, −8), (4, −6), (6, −4), (8, −3), (12, −2), (24, −1), and the sign-flipped versions.

The pair 8 and −3 sums to 8 + (−3) = 5 and multiplies to 8 · (−3) = −24.

Result: x^2 + 5x − 24 = (x + 8)(x − 3).

FOIL check:

(x + 8)(x − 3) = x^2 − 3x + 8x − 24 = x^2 + 5x − 24.

The outer and inner terms combine to the middle coefficient. If they did not, you chose the wrong pair.

A near-miss pair is 6 and −4: product −24, sum 2, which would factor x^2 + 2x − 24, a different trinomial. Another near miss is 12 and −2: sum 10, for x^2 + 10x − 24. Listing pairs first, then checking sums, is faster than guessing binomials at random.

Order of binomials does not matter: (x − 3)(x + 8) is the same factorization. Writing (x + 3)(x − 8) is not the same: that expands to x^2 − 5x − 24, which has the wrong middle sign. On a CAT, the sign-flipped middle is the most common distractor next to the correct product.

When a ≠ 1: the ac method and grouping

For ax^2 + bx + c with a ≠ 1, the leading coefficient refuses to sit quietly in one binomial. The ac method restores grouping:

1. Compute the product a · c.
2. Find integers p and q with p · q = ac and p + q = b.
3. Rewrite the middle term as px + qx.
4. Factor the resulting four-term polynomial by grouping.
5. Check by FOIL.

Worked example: 6x² − 7x − 3

Here a = 6, b = −7, c = −3, so ac = 6 · (−3) = −18.

Find two integers that multiply to −18 and add to −7.

The working pair is −9 and 2. Split the middle term:

6x^2 − 7x − 3 = 6x^2 − 9x + 2x − 3

Group:

= (6x^2 − 9x) + (2x − 3) = 3x(2x − 3) + 1(2x − 3)

The common binomial is 2x − 3:

= (3x + 1)(2x − 3)

FOIL check:

(3x + 1)(2x − 3) = 3x · 2x + 3x · (−3) + 1 · 2x + 1 · (−3)

= 6x^2 − 9x + 2x − 3 = 6x^2 − 7x − 3.

The original polynomial returns, so the factorization is correct.

You could have split as 6x^2 + 2x − 9x − 3 instead. Grouping still works:

(6x^2 + 2x) + (−9x − 3) = 2x(3x + 1) − 3(3x + 1) = (2x − 3)(3x + 1).

Same factors, opposite order. Either split is legal. What is not legal is splitting only part of −7x or dropping a sign on the second group so the binomials no longer match.

Where the binomial coefficients come from

Each binomial takes one factor of a and one factor of c in a compatible way. For 6x^2 − 7x − 3, the candidates (6x + 1)(x − 3) expand to 6x^2 − 18x + x − 3 = 6x^2 − 17x − 3, which is the wrong middle term. (3x − 1)(2x + 3) expands to 6x^2 + 9x − 2x − 3 = 6x^2 + 7x − 3, which is the sign-flipped middle. The ac method exists so you do not have to trial-and-error every factor pair of 6 against every factor pair of 3.

A GCF can appear before the ac method. 12x^2 − 14x − 6 has GCF 2: first write 2(6x^2 − 7x − 3), then factor the leftover as above to get 2(3x + 1)(2x − 3). Forgetting the opening GCF does not make the later algebra illegal, but it forces larger ac products (ac = −72 instead of −18) and more pairs to test. Section 4.1’s GCF-first habit is the cheapest way to shrink an AAF trinomial.

Prime versus composite over the integers

A polynomial with integer coefficients is composite over the integers when it factors as a product of non-constant integer-coefficient polynomials of lower degree. It is prime over the integers when no such factorization exists.

x^2 + 5x − 24 is composite: (x + 8)(x − 3). 6x^2 − 7x − 3 is composite: (3x + 1)(2x − 3). x^2 + x + 1 is prime over the integers. The discriminant is 1 − 4 = −3 < 0, so it does not even factor over the reals. No integer pair multiplies to 1 and adds to 1 except pairs that fail the sum (1 and 1 sum to 2).

x^2 + 4x + 1 is also prime over the integers even though it factors over the reals: the roots are −2 ± √3, which are not integers, and the binomials would not have integer coefficients. AAF factoring items that say “factor over the integers” or simply show integer-coefficient options are asking for integer binomials. If none exist, the honest answer is that the trinomial does not factor over the integers.

When the job is to solve ax^2 + bx + c = 0 and the trinomial is prime over the integers, do not keep hunting pairs. Use the quadratic formula or complete the square — that toolkit lives in Quadratic Formula and Completing the Square. Factoring is the wrong tool for that trinomial, not a personal failure.

A quick integer test: if you have listed every factor pair of ac and none sum to b, you are done. There is no hidden pair. For 2x^2 + 3x + 2, ac = 4 and the pairs (1, 4), (2, 2), (4, 1) and their negatives never sum to 3. The trinomial is prime over the integers. Discriminant 9 − 16 = −7 confirms there are not even real linear factors.

FOIL is the check, not the method

FOIL (First, Outer, Inner, Last) is how you multiply two binomials. It is the inverse of factoring, so it is the check:

• First: the two first terms, which rebuild ax^2.
• Outer and Inner: which combine to rebuild bx.
• Last: the two last terms, which rebuild c.

If FOIL does not reproduce the original middle coefficient, the factorization is wrong even if the First and Last pieces look right. That is the most common AAF miss on a ≠ 1 items: the ends match and the middle is off by a sign or by a factor of 2.

Checking one substituted x-value is a partial check, not a proof. (3x + 1)(2x − 3) at x = 0 gives 1 · (−3) = −3, matching c, but a wrong pair could still share the same constant. FOIL checks all three coefficients at once.

AAF habits for trinomials

1. Pull out a GCF first (Section 4.1). A leftover 2x^2 + 5x + 2 is easier than 4x^2 + 10x + 4.
2. If a = 1, list factor pairs of c and keep the pair whose sum is b.
3. If a ≠ 1, compute ac, split the middle, group.
4. Glance at the perfect-square test from Section 4.1: if |b| equals twice the product of the square roots of a and c, write a square and skip the pair list.
5. FOIL.
6. If no integer pair exists, say the trinomial is prime over the integers and, for solving, switch to the quadratic formula.

Cubics and higher-degree polynomials reuse every move in this section. Grouping, difference of squares after a GCF, and factor completely all return in Section 4.3.