Main content

## SAT

### Unit 6: Lesson 4

Passport to Advanced Math: lessons by skill- Solving quadratic equations | Lesson
- Interpreting nonlinear expressions | Lesson
- Quadratic and exponential word problems | Lesson
- Manipulating quadratic and exponential expressions | Lesson
- Radicals and rational exponents | Lesson
- Radical and rational equations | Lesson
- Operations with rational expressions | Lesson
- Operations with polynomials | Lesson
- Polynomial factors and graphs | Lesson
- Graphing quadratic functions | Lesson
- Graphing exponential functions | Lesson
- Linear and quadratic systems | Lesson
- Structure in expressions | Lesson
- Isolating quantities | Lesson
- Function Notation | Lesson

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# 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:

- Factor polynomial expressions
- 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?

### Factoring quadratic expressions

**Note:**We cover how to factor quadratic expressions in the form x, squared, plus, b, x, plus, c in detail in the

**Solving quadratic equations**lesson.

Quadratic expressions that have an x, squared-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, squared, plus, 12, x, minus, 15, we can factor out a 3 first and then factor the quadratic expression x, squared, plus, 4, x, minus, 5.

To factor a quadratic expression in the form x, squared, plus, b, x, plus, c:

- Find two numbers whose product is equal to c and whose sum is equal to b.
- The two factors of the expression are each the sum of x and one of the numbers from Step 1.

**Example:**Factor x, squared, plus, 9, x, minus, 10.

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

**factor by grouping**.### Factoring quadratics by grouping

If you'd like to review this technique, we recommend watching this video before proceeding!

### How do we factor by grouping?

The first step of factoring a, x, squared, plus, b, x, plus, c is familiar: we're looking for two integers whose product is equal to a, c and whose sum is equal to b.

For example, let's factor 2, x, squared, plus, 9, x, minus, 5.

We are looking for two numbers that meet the following criteria:

- Their product is equal to left parenthesis, 2, right parenthesis, left parenthesis, minus, 5, right parenthesis, equals, minus, 10.
- Their sum is equal to 9.

The numbers 10 and minus, 1 work:

- left parenthesis, 10, right parenthesis, left parenthesis, minus, 1, right parenthesis, equals, minus, 10
- 10, plus, left parenthesis, minus, 1, right parenthesis, equals, 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 9, x as 10, x, minus, x:Next, we

*group*the terms into two pairs:Notice that we can factor each pair of terms:

When we do this, we see that both of the initial pairs contain the factor x, plus, 5. This means we can factor out the x, plus, 5 from the overall expression:

Therefore, 2, x, squared, plus, 9, x, minus, 5, equals, left parenthesis, x, plus, 5, right parenthesis, left parenthesis, 2, x, minus, 1, right parenthesis.

**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, squared, plus, b, x, plus, c:

- Factor out any integers if possible. If this results in the product of an integer and a quadratic expression in the form x, squared, plus, b, x, plus, c, follow the steps for factoring x, squared, plus, b, x, plus, c shown above.
- Find two numbers whose product is equal to a, c and whose sum is equal to b.
- Use the two numbers from Step 1 to split b, x into two x-terms.
- Group the resulting expression into two pairs of terms: one pair should have an x, squared-term and an x-term, and the other pair should have an x-term and a constant term.
- Factor out an expression containing x from the pair with an x, squared-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.
- 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, squared, minus, 7, x, minus, 3.

### Try it!

## How do I use special factoring?

### Factor difference of squares: leading coefficient does not equal, 1

### 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, squared, check the other terms to see whether the expression satisfies the criteria for special factoring.

**Square of sum:**a, squared, plus, 2, a, b, plus, b, squared, equals, left parenthesis, a, plus, b, right parenthesis, squared

**Square of difference:**a, squared, minus, 2, a, b, plus, b, squared, equals, left parenthesis, a, minus, b, right parenthesis, squared

**Difference of squares:**a, squared, minus, b, squared, equals, left parenthesis, a, plus, b, right parenthesis, left parenthesis, a, minus, b, right parenthesis

To use special factoring to factor a polynomial expression:

- Factor out any common factors if possible.
- Recognize that one or more terms in the expression are perfect squares.
- Confirm that all of the terms in the expression satisfy the criteria for special factoring.
- Apply the appropriate special factoring rule.

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

Factor 4, x, squared, plus, 12, x, plus, 9.

Factor 9, y, start superscript, 4, end superscript, minus, 25, x, squared.

### Try it!

## Your turn!

## Things to remember

Square of sum: a, squared, plus, 2, a, b, plus, b, squared, equals, left parenthesis, a, plus, b, right parenthesis, squared

Square of difference: a, squared, minus, 2, a, b, plus, b, squared, equals, left parenthesis, a, minus, b, right parenthesis, squared

Difference of squares: a, squared, minus, b, squared, equals, left parenthesis, a, plus, b, right parenthesis, left parenthesis, a, minus, b, right parenthesis

## 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.

Thanks in advance.(3 votes)- 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?(10 votes)

- For under things to remember, do you mean step 2 instead of step 1 for number three?(0 votes)
- I agree, this looks like a typo in the article. You can tell the people that can change it by clicking the "Report a Mistake" link when you go to answer a question.(7 votes)

- 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!(5 votes)

- 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)