7.2 Factor Trinomials of the Form x2+bx+c

Factoring Trinomials of the Form $x^2+bx+c$

Factoring trinomials reverses the multiplication of two binomials, breaking a quadratic expression back into simpler pieces. This skill is essential for solving quadratic equations and simplifying algebraic expressions throughout the rest of the course.

Factoring Trinomials of the Form $x^2+bx+c$

Finding factors for $x^2+bx+c$

The core idea: you need two numbers that multiply to give you $c$ (the constant term) and add to give you $b$ (the coefficient on $x$). Once you find those two numbers, you write the factored form as $(x + \text{first number})(x + \text{second number})$.

Example: Factor $x^2+5x+6$

1. Ask: what two numbers multiply to 6 and add to 5?
2. Try factor pairs of 6: (1, 6) and (2, 3).
3. Since $2 + 3 = 5$, those are your numbers.
4. Write the answer: $(x+2)(x+3)$

You can always check your work by FOILing the result back out. $(x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$ ✓

Factoring with negative terms

Negative signs change which factor pairs you're looking for, but the same core method applies. Pay close attention to the signs of $b$ and $c$.

When $b$ is negative and $c$ is positive, both numbers must be negative. That's because two negatives multiply to a positive (giving you $c$) and add to a negative (giving you $b$).

• Factor $x^2-7x+12$: you need two numbers that multiply to $+12$ and add to $-7$.
• The pair $-3$ and $-4$ works: $(-3)(-4) = 12$ and $(-3)+(-4) = -7$.
• Factored form: $(x-3)(x-4)$

When $c$ is negative, one number is positive and the other is negative. Their product must be negative (giving you $c$), and their sum must equal $b$.

• Factor $x^2+2x-15$: you need two numbers that multiply to $-15$ and add to $+2$.
• The pair $5$ and $-3$ works: $(5)(-3) = -15$ and $5+(-3) = 2$.
• Factored form: $(x+5)(x-3)$

AC method for complex trinomials

The AC method handles trinomials where the leading coefficient isn't 1, like $ax^2+bx+c$ with $a \neq 1$. This goes beyond the $x^2+bx+c$ form, but it builds directly on the same find-two-numbers strategy.

Steps for the AC method:

1. Multiply $a \times c$ to get the product $ac$.
2. Find two numbers that multiply to $ac$ and add to $b$.
3. Rewrite the middle term ($bx$) as two separate terms using those numbers.
4. Factor by grouping (split into two pairs and factor each pair).

Example: Factor $2x^2+7x+3$

1. $a \times c = 2 \times 3 = 6$
2. Two numbers that multiply to 6 and add to 7: 1 and 6.
3. Rewrite the middle term: $2x^2+1x+6x+3$
4. Factor by grouping:
   • Group: $(2x^2+x)+(6x+3)$
   • Factor each group: $x(2x+1)+3(2x+1)$
   • Factor out the common binomial: $(2x+1)(x+3)$

The grouping step works because both groups share the same binomial factor $(2x+1)$. If your groups don't share a common binomial, double-check your number pair from step 2.