Algebra I/Quadratic Systems/Factoring Trinomials

Algebra I Regents in 22 days

Algebra I · Lesson 3

Factoring Trinomials

Breaking a quadratic expression into two binomials you can multiply back together.

Your teacher gave back a quiz and one problem said: "Expand $(x+3)(x+5)$ ." You used FOIL, got $x^2 + 8x + 15$ , and marked it correct. Now the next quiz flips it. The problem says: "Factor $x^2 + 8x + 15$ ." Same math, opposite direction. Factoring a trinomial means starting with the expanded form and working backwards to the two binomials that produced it. That reverse process is what this lesson is about.

Start with the simpler case: a trinomial where the leading coefficient is 1, meaning the expression looks like $x^2 + bx + c$ . When you FOILed $(x+3)(x+5)$ , the 8 came from $3+5$ and the 15 came from $3 \cdot 5$ . That pattern is the key. To factor $x^2 + bx + c$ , you need two numbers that multiply to $c$ and add to $b$ .

x^2 + bx + c = (x + p)(x + q) \quad \text, p \cdot q = c \text, p + q = b

Take $x^2 + 7x + 12$ . You need two numbers whose product is 12 and whose sum is 7. List the factor pairs of 12: $1 \times 12$ , $2 \times 6$ , $3 \times 4$ . Check which pair adds to 7. That is 3 and 4. So the factored form is $(x+3)(x+4)$ .

Factor x² + 7x + 12

p \cdot q = 12, \quad p + q = 7

Write down the two conditions your numbers have to satisfy.

1 \times 12 = 12, \quad 1 + 12 = 13

Too big for the sum — move on.

2 \times 6 = 12, \quad 2 + 6 = 8

Still not 7 — keep going.

3 \times 4 = 12, \quad 3 + 4 = 7 \checkmark

Both conditions satisfied. These are your numbers.

x^2 + 7x + 12 = (x+3)(x+4)

Write the factored form using 3 and 4.

Negatives come into play when $c$ is negative or when $b$ is negative. If $c$ is negative, one of your two numbers must be positive and the other negative — that is the only way a product is negative. If $c$ is positive but $b$ is negative, both numbers are negative.

Factor x² - 3x - 10

p \cdot q = -10, \quad p + q = -3

The product is negative, so one number is positive and one is negative.

(-5)(2) = -10, \quad -5 + 2 = -3 \checkmark

Think about which sign needs to be on the bigger number. Since the sum is negative, the negative number needs to be the larger one in absolute value.

x^2 - 3x - 10 = (x-5)(x+2)

Slot the numbers in. The negative goes with 5, the positive with 2.

You can always verify a factored answer by expanding it with FOIL. If you get back to the original trinomial, you are correct.

Now for the harder case: $ax^2 + bx + c$ where $a \neq 1$ . The simple find-two-numbers trick no longer works directly because the leading coefficient changes the FOIL pattern. The method that works cleanly here is called the AC method.

Multiply $a$ and $c$ together — that product is called AC. Find two numbers that multiply to AC and add to $b$ . Then rewrite the middle term using those two numbers, splitting the trinomial into four terms. Factor by grouping. The structure becomes clear in a worked example.

Factor 2x² + 7x + 3

a = 2, \quad b = 7, \quad c = 3

Identify the pieces before doing anything else.

AC = 2 \cdot 3 = 6

Multiply the leading coefficient by the constant.

p \cdot q = 6, \quad p + q = 7

Now find two numbers that multiply to 6 and add to 7.

1 \times 6 = 6, \quad 1 + 6 = 7 \checkmark

Found them: 1 and 6.

2x^2 + 1x + 6x + 3

Rewrite the middle term 7x as 1x + 6x. The order matters — group the first two and last two.

x(2x + 1) + 3(2x + 1)

Factor an x out of the first pair, and a 3 out of the second pair. If the binomials in parentheses don't match, go back and check your split.

(x + 3)(2x + 1)

The shared binomial (2x+1) factors out, leaving (x+3) behind.

Factor 6x² - 11x + 4

a = 6, \quad b = -11, \quad c = 4

Identify each coefficient, including the negative sign on b.

AC = 6 \cdot 4 = 24

Multiply a and c. Both are positive, so AC is positive.

p \cdot q = 24, \quad p + q = -11

Product is positive, sum is negative — both numbers must be negative.

(-3)(-8) = 24, \quad -3 + (-8) = -11 \checkmark

Both negative, product 24, sum -11. That works.

6x^2 - 3x - 8x + 4

Rewrite -11x as -3x + (-8x). Split and group.

3x(2x - 1) - 4(2x - 1)

Factor 3x from the first pair. Factor -4 from the second pair — the negative sign is critical to make the binomials match.

(3x - 4)(2x - 1)

The shared binomial (2x-1) factors out.

A trinomial that cannot be factored using integers is called a prime trinomial. If you list all factor pairs of AC and none of them add to $b$ , the trinomial is prime over the integers.

Practice Questions

x^2 + 9x + 20

x^2 - 2x - 15

3x^2 + 10x - 8

Regents Corner

On Part II and Part III of the Algebra I Regents, factoring trinomials often appears as a step inside a larger problem — solving a quadratic equation by factoring, or identifying zeros of a function. You must show the factored form and then set each factor equal to zero. Writing only the factored form without solving for $x$ will cost you points. Writing only the answers without the factored form will also cost you points.

When using the AC method, students sometimes split the middle term correctly but then factor by grouping carelessly and end up with two different binomials in parentheses — for example, getting x(3x - 2) + 4(3x + 2). If the two binomials do not match exactly, sign and all, the grouping is wrong. Go back and check whether you factored out a negative correctly from the second group.

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Factoring — GCF and Difference of Squares

Operations with Radicals