Factoring Quadratic Expressions
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Always look for a common factor first. Then, for $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. When the leading coefficient is not $1$, split the middle term and factor by grouping.
Two patterns are worth knowing by sight: the difference of squares, $a^2 - b^2 = (a - b)(a + b)$, and perfect squares such as $x^2 + 10x + 25 = (x + 5)^2$.
On the SAT you are usually asked to choose an equivalent factored form, to find a factor, or to find a missing constant when two forms are equal for all $x$. Whatever the question, you can check a factored answer by multiplying it back out, and I encourage students to do that whenever they have the time.
Four examples
Example 1: a negative constant
Factor $x^2 - 5x - 24$.
- You need a product of $-24$ and a sum of $-5$. The product is negative, so the signs are opposite.
- The pairs for $24$ are $1 \cdot 24$, $2 \cdot 12$, $3 \cdot 8$, $4 \cdot 6$. The pair $-8$ and $3$ sums to $-5$.
- $x^2 - 5x - 24 = (x - 8)(x + 3)$. Check: $x^2 + 3x - 8x - 24 = x^2 - 5x - 24$.
Example 2: factoring by grouping
Factor $6x^2 + x - 2$.
- Multiply $a \cdot c = 6 \cdot (-2) = -12$. Find two numbers with product $-12$ and sum $1$: $4$ and $-3$.
- Split the middle term: $6x^2 + 4x - 3x - 2$.
- Group: $2x(3x + 2) - 1(3x + 2)$.
- Factor out $3x + 2$: $(2x - 1)(3x + 2)$.
Example 3: common factor, then a pattern
Factor $2x^2 - 50$ completely.
- Take out the common factor $2$: $2(x^2 - 25)$.
- $x^2 - 25$ is a difference of squares: $(x - 5)(x + 5)$.
- $2x^2 - 50 = 2(x - 5)(x + 5)$.
Example 4: finding a missing constant
If $x^2 + kx - 12 = (x - 2)(x + m)$ for all values of $x$, what is $k$?
- Expand the right side: $x^2 + mx - 2x - 2m = x^2 + (m - 2)x - 2m$.
- Match the constants: $-2m = -12$, so $m = 6$.
- Match the $x$ terms: $k = m - 2 = 4$.
What goes wrong
Skipping the common factor. $2x^2 - 50$ is not fully factored until the $2$ comes out.
Getting the signs backwards. $(x + 8)(x - 3)$ multiplies to $x^2 + 5x - 24$, not $x^2 - 5x - 24$. Multiplying back out will catch this.
Factoring a sum of squares. $x^2 + 9$ does not factor over the real numbers. Only a difference of squares does.
Confusing factors and zeros. The factor $(x - 3)$ means the zero is $x = 3$, not $x = -3$.
Forgetting the leading coefficient. $3x^2 - 11x - 4$ is not $(x - 4)(x + 1)$. The $3$ has to appear in one of the factors.
Quick reference
| Form | Method | Example |
|---|---|---|
| Common factor | Take out the greatest common factor | $3x^2 + 12x = 3x(x + 4)$ |
| $x^2 + bx + c$ | Two numbers with product $c$ and sum $b$ | $x^2 + 7x + 12 = (x + 3)(x + 4)$ |
| $ax^2 + bx + c$ | Two numbers with product $ac$ and sum $b$, then group | $2x^2 + 7x + 3 = (2x + 1)(x + 3)$ |
| Difference of squares | $a^2 - b^2 = (a - b)(a + b)$ | $x^2 - 49 = (x - 7)(x + 7)$ |
| Perfect square | $a^2 + 2ab + b^2 = (a + b)^2$ | $x^2 + 10x + 25 = (x + 5)^2$ |
A sign tip: if $c$ is positive, both numbers have the same sign as $b$. If $c$ is negative, the numbers have opposite signs, and the larger one carries the sign of $b$.
Your turn
Factor each one on paper first. If you are not sure, multiply your answer back out before you check the solution.
- A$(x + 2)(x + 6)$
- B$(x + 3)(x + 4)$
- C$(x + 1)(x + 12)$
- D$(x - 3)(x - 4)$
Show solution
- A$(x - 7)^2$
- B$(x + 7)^2$
- C$(x - 7)(x + 7)$
- D$(x - 49)(x + 1)$
Show solution
- A$(x + 8)(x - 3)$
- B$(x - 6)(x + 4)$
- C$(x - 12)(x + 2)$
- D$(x - 8)(x + 3)$
Show solution
- A$(2x - 1)(3x + 2)$
- B$(2x + 1)(3x - 2)$
- C$(6x - 1)(x + 2)$
- D$(3x - 1)(2x + 2)$
Show solution
- A$2(x - 5)^2$
- B$2(x - 5)(x + 5)$
- C$(2x - 25)(x + 2)$
- D$2(x^2 - 25)$
Show solution
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- A$x + 4$
- B$3x - 1$
- C$3x + 4$
- D$x - 4$
Show solution
- A$(3x - 2)^2$
- B$(3x + 2)^2$
- C$(9x - 2)(x - 2)$
- D$(3x - 4)(3x - 1)$
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