If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Structure in expressions | Lesson

## What are "Structure in expressions" questions, and how frequently do they appear on the test?

Structure in expressions questions require you to understand how to factor polynomials.
In this lesson, we'll learn to:
1. Factor polynomial expressions
2. Use knowledge of factoring to evaluate expressions
On your official SAT, you'll likely see 1 to 2 questions that test your ability to see structure in expressions. Other questions may also require you to factor quadratic or polynomial expressions as a step toward solving or graphing equations.
You can learn anything. Let's do this!

## How do I factor quadratic expressions?

Note: We cover how to factor quadratic expressions in the form ${x}^{2}+bx+c$ in detail in the Solving quadratic equations lesson.
Quadratic expressions that have an ${x}^{2}$-coefficient that is not $1$ are more difficult to factor. We should try to factor out any common factors if possible.
For example, in the expression $3{x}^{2}+12x-15$, we can factor out a $3$ first and then factor the quadratic expression ${x}^{2}+4x-5$.
To factor a quadratic expression in the form ${x}^{2}+bx+c$:
1. Find two numbers whose product is equal to $c$ and whose sum is equal to $b$.
2. The two factors of the expression are each the sum of $x$ and one of the numbers from Step 1.

Example: Factor ${x}^{2}+9x-10$.

When we can't factor out an integer, we can use a technique called factor by grouping.

If you'd like to review this technique, we recommend watching this video before proceeding!
Factoring quadratics by groupingSee video transcript

### How do we factor by grouping?

The first step of factoring $a{x}^{2}+bx+c$ is familiar: we're looking for two integers whose product is equal to $ac$ and whose sum is equal to $b$.
For example, let's factor $2{x}^{2}+9x-5$.
We are looking for two numbers that meet the following criteria:
• Their product is equal to $\left(2\right)\left(-5\right)=-10$.
• Their sum is equal to $9$.
The numbers $10$ and $-1$ work:
• $\left(10\right)\left(-1\right)=-10$
• $10+\left(-1\right)=9$
These two numbers alone do not give us the factors. Instead, they tell us how to split up the $x$-term of the expression. We can rewrite $9x$ as $10x-x$:
$2{x}^{2}+9x-5=2{x}^{2}+10x-x-5$
Next, we group the terms into two pairs:
$2{x}^{2}+10-x-5$
Notice that we can factor each pair of terms:
$\begin{array}{rl}& 2{x}^{2}+10x-x-5\\ \\ & =2x\left(x+5\right)-1\left(x+5\right)\end{array}$
When we do this, we see that both of the initial pairs contain the factor $x+5$. This means we can factor out the $x+5$ from the overall expression:
$\begin{array}{rl}& 2x\left(x+5\right)-1\left(x+5\right)\\ \\ & =x+5\cdot \left(\frac{2x\left(x+5\right)}{x+5}-\frac{1\left(x+5\right)}{x+5}\right)\\ \\ & =\left(x+5\right)\left(2x-1\right)\end{array}$
Therefore, $2{x}^{2}+9x-5=\left(x+5\right)\left(2x-1\right)$.
Note: Factoring by grouping can be difficult, and there are often other strategies that can get you to the answer on test day. For example, you might be able to simply test multiple choice options by using FOIL, or plug in simple values for $x$ to find a match!
To factor a quadratic expression in the form $a{x}^{2}+bx+c$:
1. Factor out any integers if possible. If this results in the product of an integer and a quadratic expression in the form ${x}^{2}+bx+c$, follow the steps for factoring ${x}^{2}+bx+c$ shown above.
2. Find two numbers whose product is equal to $ac$ and whose sum is equal to $b$.
3. Use the two numbers from Step 1 to split $bx$ into two $x$-terms.
4. Group the resulting expression into two pairs of terms: one pair should have an ${x}^{2}$-term and an $x$-term, and the other pair should have an $x$-term and a constant term.
5. Factor out an expression containing $x$ from the pair with an ${x}^{2}$-term and an $x$-term. Factor out a constant from the pair with an $x$-term and a constant term. These two pairs should now share a binomial factor.
6. The shared binomial factor is one factor of the quadratic expression. The expression containing $x$ and the constant factored out in Step 4 combine to form the other factor of the quadratic expression.

Example: Factor $6{x}^{2}-7x-3$.

### Try it!

$4{x}^{2}+25x+6$
To factor the quadratic equation above, first we need to find two numbers whose product is equal to the product of the ${x}^{2}$-coefficient and the constant term, $4\cdot 6=24$, and whose sum is equal to the $x$-coefficient, $25$.
These two numbers are
.
Next, we use the two numbers to rewrite $4{x}^{2}+25x+6$ as $4{x}^{2}+24x+x+6$.
The greatest common factor of $4{x}^{2}$ and $24x$ is
.
This means we can rewrite $4{x}^{2}+24x+x+6$ as:
$4x\left(x+6\right)+1\left(x+6\right)$
In the form of the quadratic expression above, $x+6$ is a common factor. This $x+6$ is a factor of $4{x}^{2}+25+6$, and the other factor is
.

## How do I use special factoring?

### Factor difference of squares: leading coefficient $\ne 1$‍

Factoring difference of squares: leading coefficient ≠ 1See video transcript

### Special factoring

Special factoring rules are shortcuts for factoring polynomials with specific combinations of terms. Many students find that when they recognize opportunities to apply special factoring rules, they save valuable time on test day.
Anytime we see multiple perfect square terms in a polynomial expression, e.g., $9$ or $4{x}^{2}$, check the other terms to see whether the expression satisfies the criteria for special factoring.
Square of sum: ${a}^{2}+2ab+{b}^{2}=\left(a+b{\right)}^{2}$
Square of difference: ${a}^{2}-2ab+{b}^{2}=\left(a-b{\right)}^{2}$
Difference of squares: ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$
To use special factoring to factor a polynomial expression:
1. Factor out any common factors if possible.
2. Recognize that one or more terms in the expression are perfect squares.
3. Confirm that all of the terms in the expression satisfy the criteria for special factoring.
4. Apply the appropriate special factoring rule.

#### Let's look at some examples!

Factor $4{x}^{2}+12x+9$.

Factor $9{y}^{4}-25{x}^{2}$.

### Try it!

Try: use special factoring to evaluate an expression
If $x+y=8$ and $x-y=2$, what is the value of $\left({x}^{2}-{y}^{2}\right)\left({x}^{2}-2xy+{y}^{2}\right)$ ?
${x}^{2}-{y}^{2}$ is a difference of squares. This means we can factor the expression into
.
${x}^{2}-2xy+{y}^{2}$ can be factored into a square of difference, or
.
Since $x+y=8$ and $x-y=2$, we can plug in $8$ for each $x+y$ and $2$ for each $x-y$. This means the value of $\left({x}^{2}-{y}^{2}\right)\left({x}^{2}-2xy+{y}^{2}\right)$ is equal to
, or
.

If $8{x}^{2}-5x-22=\left(ax+11\right)\left(bx-2\right)$, where $a$ and $b$ are constants, what is the value of $a-b$ ?

Practice: use special factoring
Which of the following is equivalent to $4{x}^{2}+28x+49$ ?

Practice: use special factoring
Which of the following is equivalent to $2{x}^{5}-8x$ ?

## Things to remember

Square of sum: ${a}^{2}+2ab+{b}^{2}=\left(a+b{\right)}^{2}$
Square of difference: ${a}^{2}-2ab+{b}^{2}=\left(a-b{\right)}^{2}$
Difference of squares: ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$

## Want to join the conversation?

• In the factorization of 8x^2-5x-22 above, how did u write ab=8 and a(-2)+11.b= -5 ? I didn't get that clearly. Requesting to explain once again. • We can write that ab = 8 because in the right side of the equation, both a and b are multiplied by x. When you multiply each factor in the right side together, you'd end up getting abx^2 as one of the terms. Since this is the only way that you can get x^2 from the right side and the equation is true for all x values, we're allowed to say that 8x^2 has to be equal to abx^2, which makes ab = 8.
We get -2a + 11b = 5 in a similar manner. The two ways of making an x term with the right side are when you multiply a by -2 when you multiply the factors together or b by 11. Put them together and you get -2ax + 11bx = 5x, or -2a + 11b = 5.
From there you solve the system of equations for a and b, and then use that to finish off the question. Did this clear it up a bit?
• For under things to remember, do you mean step 2 instead of step 1 for number three?
(1 vote) • In the step in factor by grouping where we go from 2x^2+9x-5=2x^2+10x-x-5 to 2x+10−x−5 what do we do to get this. What happens to the other 2x^2, 9x and other 5?
(1 vote) • This is basically what the process looks like:
We want to factor 2x^2 + 9x - 5. To do that, we split the middle term into two numbers that allow us to take a common factor out of two terms, which'll allow us to create factors. To do this, we want the two numbers to multiply to a*c and sum to b, where a, b, and c are the coefficients in degree order. This comes out to be 10 and -1.
= 2x^2 + 10x - x - 5
Now, we factor a 2x out of the first two terms:
= 2x (x + 5) + (-x - 5)
From here, we want our thing inside the parentheses to be the same for both terms, so we factor a -1 out of the 2nd term:
= 2x (x + 5) - (x + 5)
Since we have two terms being multiplied by the same thing, we can apply the distributive property in reverse to condense this into two factors:
= (2x - 1)(x + 5)
The 2x^2, 9x, and 5 were just transformed along the way to become these two factors. If you multiply them together you'll se that you get 2x^2 + 9x + 5 once again.
In the step you're talking about, there's a typo; it should be "2x^2 + 10x + x - 5". This is the exact same as the right side of the equation in the previous step and it all checks out.
Hope this helps!
• How come this section never taught us the formula for x^3 + b^3 = (x+b)(x^2 - xb + b^2). When I came upon the practice showing me this problem, I was really stumped.
(1 vote) • Hi! I was doing this question in the SAT prep section.
(yt)^6t -(lv)^8v Which of the following is equivalent to the expression above?

y^8vl^8v-t^6ty^6t
((lv)^4v+(yt)^3t)((lv)^4v-(yt)^3t
(vl^v4+ty^t3)(ty^t3-vl^v4

The explanation says the final two choices are written in a funny form: ((lv)^4v+(yt)^3t)((lv)^4v-(yt)^3t
The terms have been halved! The difference of squares formula allows us to halve powers in this way.
a=yt^3t b=lv^4v
a^2-b^2=(a+b)(a+b)

My question is why do we use this formula and why do we rule out the first answer choice?
(1 vote) 