Lesson 6.2: Factor Trinomials

New Concept

Factoring trinomials is like reverse-multiplication. You'll learn how to break down expressions like $ax^2 + bx + c$ into their original binomial factors. We'll explore strategies for different trinomial forms, including trial and error, the 'ac' method, and substitution.

What’s next

Now, let's master this skill. You'll work through guided examples for each method, followed by practice cards and challenge problems to test your factoring fluency.

Property

To factor a trinomial of the form $x^2 + bx + c$, we need two factors $(x + m)$ and $(x + n)$ where the two numbers $m$ and $n$ multiply to $c$ and add to $b$.

How to factor trinomials of the form $x^2 + bx + c$

1. Write the factors as two binomials with first terms $x$: $(x \quad)(x \quad)$.
2. Find two numbers $m$ and $n$ that multiply to $c$, $m \cdot n = c$, and add to $b$, $m + n = b$.
3. Use $m$ and $n$ as the last terms of the factors.
4. Check by multiplying the factors.

Strategy for Determining Signs
When $c$ is positive, $m$ and $n$ have the same sign as $b$.

• If $b$ is positive, $m$ and $n$ are positive. Example: $x^2 + 5x + 6 = (x+2)(x+3)$.
• If $b$ is negative, $m$ and $n$ are negative. Example: $x^2 - 6x + 8 = (x-4)(x-2)$.

Property

How to factor trinomials of the form $ax^2 + bx + c$ using trial and error.

1. Write the trinomial in descending order of degrees.
2. Factor any GCF. If the leading coefficient is negative, the GCF will be negative.
3. Find all the factor pairs of the first term ($ax^2$). These will be the first terms of your binomial factors.
4. Find all the factor pairs of the third term ($c$). These will be the last terms.
5. Test all possible combinations of the factors. The correct combination is the one where the sum of the inner and outer products equals the middle term ($bx$).
6. Check by multiplying.

Examples

• To factor $2x^2 + 9x + 10$, the factors of $2x^2$ are $(x, 2x)$. The factors of 10 are $(1, 10)$ and $(2, 5)$. Testing combinations, we find $(x+2)(2x+5)$ gives $5x + 4x = 9x$. So, the factors are $(x+2)(2x+5)$.
• To factor $6y^2 - 19y + 10$, we need two negative factors for 10. Factors of $6y^2$ include $(2y, 3y)$. Testing combinations, $(2y-5)(3y-2)$ gives $-4y - 15y = -19y$. So, the factors are $(2y-5)(3y-2)$.
• To factor $12x^3 + 4x^2 - 16x$, first factor out the GCF, $4x$. This gives $4x(3x^2 + x - 4)$. Factoring the trinomial, we get $4x(3x+4)(x-1).

Explanation

This method is a systematic puzzle. You list all possible factors for the first ($a$) and last ($c$) coefficients and test pairs. Keep trying combinations until the 'Outer' and 'Inner' products from FOIL add up to the middle term.

Property

How to factor trinomials of the form $ax^2 + bx + c$ using the “ac” method.

1. Factor any GCF.
2. Find the product $ac$.
3. Find two numbers, $m$ and $n$, that multiply to $ac$ ($m \cdot n = ac$) and add to $b$ ($m + n = b$).
4. Split the middle term using $m$ and $n$: $ax^2 + mx + nx + c$.
5. Factor the resulting four-term polynomial by grouping.
6. Check by multiplying the factors.

Examples

• To factor $8x^2 + 10x + 3$ using the 'ac' method, find $ac = 8 \cdot 3 = 24$. Two numbers that multiply to 24 and add to 10 are 4 and 6. Rewrite as $8x^2 + 4x + 6x + 3$. Factor by grouping: $4x(2x+1) + 3(2x+1) = (4x+3)(2x+1)$.
• To factor $12y^2 - 5y - 2$, find $ac = 12 \cdot (-2) = -24$. Two numbers that multiply to -24 and add to -5 are -8 and 3. Rewrite as $12y^2 - 8y + 3y - 2$. Factor by grouping: $4y(3y-2) + 1(3y-2) = (4y+1)(3y-2)$.
• To factor $18z^2 - 33z + 9$, first factor out the GCF, 3, to get $3(6z^2 - 11z + 3)$. For the trinomial, $ac = 6 \cdot 3 = 18$. Numbers are -2 and -9. Rewrite as $3(6z^2 - 2z - 9z + 3)$. Grouping gives $3(2z-3)(3z-1).

Explanation

This structured method avoids guesswork. You use the product 'ac' to find two numbers that let you split the middle term into two parts. This transforms the trinomial into a four-term expression that you can then factor by grouping.

Property

Sometimes a trinomial does not appear to be in the $ax^2 + bx + c$ form. However, we can often make a thoughtful substitution that will allow us to make it fit the form. It is standard to use $u$ for the substitution. In the $ax^2 + bx + c$ form, the middle term has a variable, $x$, and its square, $x^2$, is the variable part of the first term. Look for this relationship when finding a substitution.

Examples

• To factor $x^4 + 2x^2 - 15$, notice the pattern where the first term's variable part ($x^4$) is the square of the middle term's ($x^2$). Let $u=x^2$. The expression becomes $u^2 + 2u - 15$, which factors to $(u+5)(u-3)$. Substitute back: $(x^2+5)(x^2-3)$.
• To factor $(x+3)^2 + 9(x+3) + 20$, let $u = x+3$. The expression becomes $u^2 + 9u + 20$. This factors to $(u+4)(u+5)$. Substitute back: $( (x+3)+4 )( (x+3)+5 )$, which simplifies to $(x+7)(x+8)$.
• To factor $y^4 - 8y^2 + 16$, let $u = y^2$. The expression becomes $u^2 - 8u + 16$, which is a perfect square trinomial, $(u-4)^2$. Substitute back to get $(y^2-4)^2$. This can be factored further as $((y-2)(y+2))^2$ or $(y-2)^2(y+2)^2.

Explanation

When a trinomial looks too complex, like having a fourth power, simplify it with a disguise! Let a new variable, $u$, stand in for the repeated part (like $x^2$). Factor the new, simpler trinomial, then substitute the original expression back.