Distribute, combine, and rewrite expressions

Lesson progressPractice problems 0/3
Difficulty
Beginner
Estimated time
30 minutes
Techniques
Equivalent-expressionsDistributive-propertyLike-termsPolynomial-operationsGraph-overlap

What you’ll learn

  1. Spot when an SAT question wants an equivalent expression.
  2. Distribute a factor to every term inside parentheses, even when the factor is negative.
  3. Add, subtract, and multiply polynomials without losing a term or a sign.
  4. Decide when to graph the choices with full-curve overlap and when to do exact algebra.

Prerequisites

You’re ready. No earlier Aniko lesson is required.

Why this matters on the SAT

Turn layers of parentheses into one clean form

Equivalent-expression questions come in two kinds. Some give you answer choices and ask which one matches, so your job is to identify a form. Others ask you to create one: write the new form yourself, or find one of its coefficients. Which kind you're facing decides your fastest method. Here's the first kind.

SAT example

Which expression is equivalent to

(4x−1)(x+3)−(x−2)(2x+5)?(4x-1)(x+3)-(x-2)(2x+5)?
  1. A

    2x2+10x−72x^2+10x-7

  2. B

    2x2+10x+72x^2+10x+7

  3. C

    2x2+12x+72x^2+12x+7

  4. D

    6x2+12x−136x^2+12x-13

Solution to the example

Expanding two products and a subtraction by hand gives a sign lots of chances to slip. With one variable and four complete choices, you can let Desmos compare them instead. Type the original expression on line 1 and choice A on line 2. Then replace choice A with choice B. B's curve lands exactly on top of the original, everywhere on the screen, so the answer is B.

That check is called full-curve overlap. The right choice doesn't only cross the original at some point. It matches it along the whole curve.

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Line 1 is the original, and line 2 is choice B. Turn line 2 off and on. The curve stays put, because both lines draw the same parabola.

Recognize the rewrite job

Two expressions are equivalent when they give the same value for every allowed value of their variables. For example,

3(x+4)=3x+12.3(x+4)=3x+12.

The two sides look different, but distributing the 33 turns the left side into the right side, so they give the same output for every xx.

You'll recognize this kind of question from wording like:

  • Which expression is equivalent to ...?
  • The expression can be written in the form ...
  • What is the coefficient of ... after rewriting?

None of these asks you to find a value of xx. You're changing the form, or picking the matching form, and xx stays a variable the whole time. If the question gives you an equation and asks you to find an unknown, that's solving, and you balance both sides as in Solve linear equations.

Choose Desmos or algebra

Start with full-curve overlap when the question looks like the opening example:

  • The expression has only one variable, like xx.
  • The answer choices are complete expressions, so each one can go straight into Desmos.
  • Expanding by hand would be long, with several products or minus signs to track.

Here's how it goes. Type the original on line 1. Type one choice on line 2, then replace it with each of the other choices in turn. The right choice sits exactly on top of the original across the whole visible graph.

Why isn't one shared point enough? Two different expressions can give the same output at a single input. For example, x2x^2 and 2x2x both equal 44 when x=2x=2, but at x=3x=3 they give 99 and 66. So a curve that only crosses the original proves nothing. You need the whole curve to match.

Start by hand in these cases instead:

  • One short distribution settles it, like turning 2(x+7)2(x+7) into 2x+142x+14.
  • More than one variable remains, like in 4(a−b)+a4(a-b)+a.
  • The question asks you to produce a coefficient or write the form yourself, like "What is the value of cc?"
  • The expression has a variable in a denominator, like x2−4x−2\frac{x^2-4}{x-2}, where xx can't be 22. A graph can't show a restriction like that exactly.

Related: For more practice comparing expressions by graph, see Equivalent expressions by graph overlap.

Check your understanding:

Which method would you start with for each of these, and why? First, simplify −4(3x−2)-4(3x-2). Second, pick which of four long choices matches a one-variable expression built from several products.

Distribute to every term, then combine

The distributive property says that a factor outside parentheses multiplies every term inside:

a(b+c)=ab+ac.a(b+c)=ab+ac.

It works the same way with subtraction:

a(b−c)=ab−ac.a(b-c)=ab-ac.

A minus sign in front of parentheses works like a factor of −1-1. Writing the −1-1 in makes it clear that every sign inside flips:

−(2x2−5x+7)=(−1)(2x2−5x+7)=−2x2+5x−7.\begin{aligned} -(2x^2-5x+7) &=(-1)(2x^2-5x+7)\\[1.4em] &=-2x^2+5x-7. \end{aligned}

Once the parentheses are gone, combine like terms, terms with exactly the same variable part. 4x24x^2 and −x2-x^2 are like terms. 4x24x^2 and 4x4x aren't, because the powers of xx are different.

Here are both steps together:

3(2x2−x+4)+2x2+5x−1=6x2−3x+12+2x2+5x−1=(6x2+2x2)+(−3x+5x)+(12−1)=8x2+2x+11.\begin{aligned} 3(2x^2-x+4)+2x^2+5x-1 &=6x^2-3x+12+2x^2+5x-1\\[1.4em] &=(6x^2+2x^2)+(-3x+5x)+(12-1)\\[1.4em] &=8x^2+2x+11. \end{aligned}

You won't always need the grouping line in the middle, but it lays out the three like-term groups so you can check each one.

Check your understanding:

Rewrite −3(2x2−x+4)-3(2x^2-x+4). Why does the xx-term come out positive?

Common mistake:

Turning 3x2+4x3x^2+4x into 7x27x^2 combines unlike terms. One has x2x^2 and the other has xx, so their variable parts don’t match, and 3x2+4x3x^2+4x is already as simple as it gets. Plugging in x=2x=2 shows the problem: the original is 2020, but 7x27x^2 is 2828.

Add, subtract, and multiply polynomials

A polynomial is a sum of terms whose exponents are whole numbers, like 4x2−3x+84x^2-3x+8. Adding, subtracting, and multiplying polynomials uses the same two moves over and over: distribute completely, then combine like terms.

Add or subtract

When you add, the parentheses can come straight off. Then combine the matching terms:

(4x2−3x+8)+(x2+5x−11)=4x2−3x+8+x2+5x−11=5x2+2x−3.\begin{aligned} (4x^2-3x+8)+(x^2+5x-11) &=4x^2-3x+8+x^2+5x-11\\[1.4em] &=5x^2+2x-3. \end{aligned}

Subtraction is where signs go missing. The minus sign applies to the whole second polynomial, so every one of its terms flips:

(4x2−3x+8)−(x2+5x−11)=4x2−3x+8−x2−5x+11=3x2−8x+19.\begin{aligned} (4x^2-3x+8)-(x^2+5x-11) &=4x^2-3x+8-x^2-5x+11\\[1.4em] &=3x^2-8x+19. \end{aligned}

Look at the last term. Subtracting −11-11 gives +11+11.

Multiply

When you multiply two polynomials, every term in one has to multiply every term in the other. Two binomials give four products:

(2x−3)(x+4)=(2x)(x)+(2x)(4)+(−3)(x)+(−3)(4)=2x2+8x−3x−12=2x2+5x−12.\begin{aligned} (2x-3)(x+4) &=(2x)(x)+(2x)(4)+(-3)(x)+(-3)(4)\\[1.4em] &=2x^2+8x-3x-12\\[1.4em] &=2x^2+5x-12. \end{aligned}

Write out all four products before you combine anything. That's what keeps the middle term, here 8x−3x=5x8x-3x=5x, from going missing. The same every-term rule works for bigger polynomials too.

Try it yourself:

Rewrite (5x2+x−6)−(2x2−4x+3)(5x^2+x-6)-(2x^2-4x+3). Keep the minus sign in view while you take off the second set of parentheses, then group the like terms.

Check your understanding:

What did you get for the Try it yourself problem, and which signs changed?

Example: Control every product and sign

Worked example

Which expression is equivalent to

(3x+2)(x−5)−(x+1)2+4?(3x+2)(x-5)-(x+1)^2+4?
  1. A

    2x2−15x−72x^2-15x-7

  2. B

    2x2−11x−72x^2-11x-7

  3. C

    4x2−11x−74x^2-11x-7

  4. D

    2x2−15x−152x^2-15x-15

Step 1

Start with full-curve overlap

This one-variable multiple-choice question has a product, a square, and a subtraction to expand, so start in Desmos. Graph the original on line 1 and try choices A through D on line 2. Only choice A stays on top of the original the whole way:

y=(3x+2)(x−5)−(x+1)2+4y=(3x+2)(x-5)-(x+1)^2+4

and

y=2x2−15x−7.y=2x^2-15x-7.

So the answer is A, and you didn't have to do the long expansion to find it.

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The original and choice A overlap. Type another choice into line 2 and watch the curves pull apart.

Step 2

Confirm the first product

Now let's confirm it by hand. The algebra also shows where a wrong choice like B comes from. Multiply each term in 3x+23x+2 by each term in x−5x-5:

(3x+2)(x−5)=3x2−15x+2x−10=3x2−13x−10.\begin{aligned} (3x+2)(x-5) &=3x^2-15x+2x-10\\[1.4em] &=3x^2-13x-10. \end{aligned}

Step 3

Expand the square and subtract it

A square is a product of two copies, so it has a middle term: (x+1)2(x+1)^2 means (x+1)(x+1)=x2+2x+1(x+1)(x+1)=x^2+2x+1, not x2+1x^2+1. Put both expanded products in, and keep the square in parentheses until you've handled the minus sign in front of it:

(3x2−13x−10)−(x2+2x+1)+4=3x2−13x−10−x2−2x−1+4.\begin{aligned} &(3x^2-13x-10)-(x^2+2x+1)+4\\[1.4em] &=3x^2-13x-10-x^2-2x-1+4. \end{aligned}

Step 4

Combine like terms

Group the x2x^2-terms, the xx-terms, and the constants:

(3x2−x2)+(−13x−2x)+(−10−1+4)=2x2−15x−7.\begin{aligned} (3x^2-x^2)+(-13x-2x)+(-10-1+4) &=2x^2-15x-7. \end{aligned}

The algebra agrees with the graph. The answer is A.

Common mistake:

Choice B comes from writing −(x2+2x+1)-(x^2+2x+1) as −x2+2x−1-x^2+2x-1, flipping the first and last signs but not the middle one. In Desmos, B’s curve pulls away from the original, so overlap rules it out. By hand, remember that the minus sign multiplies the whole square by −1-1, so all three signs flip: −(x2+2x+1)=−x2−2x−1-(x^2+2x+1)=-x^2-2x-1.

Practice problems

The problems below don't all call for the same first move. Before you start one, decide whether Desmos or hand algebra should go first.

Distribute two linear factors

Practice problem

Which expression is equivalent to

5(2r−3)−4(r+1)?5(2r-3)-4(r+1)?
Answer choices
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Do this one by hand, since two short distributions settle it. If you’d like a check afterward, graph the original and your answer to see that they overlap.

Subtract one polynomial product from another

Practice problem

Which expression is equivalent to

(3x−2)(x+4)−(x+3)(x−5)?(3x-2)(x+4)-(x+3)(x-5)?
Answer choices
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Start with full-curve overlap here, since this is a long one-variable multiple-choice question.

Recover a coefficient after cubic cancellation

Practice problem

The expression

(2x2−3x+4)(x−5)−(x2+2x−1)(2x+3)(2x^2-3x+4)(x-5)-(x^2+2x-1)(2x+3)

is equivalent to bx2+cx+dbx^2+cx+d, where bb, cc, and dd are constants. What is the value of cc?

Calculator loads as you approach
Work this one by hand, since you need an exact coefficient. Afterward, you can graph the original and your answer to check that they overlap, but read cc from your algebra.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

  • Equivalent expressions give the same value for every allowed input, even when they look different.
  • A factor outside parentheses multiplies every term inside. A minus sign in front works like −1-1 and flips every sign.
  • Combine only like terms, the ones with exactly the same variable part.
  • Add, subtract, and multiply one layer at a time, and write out every product before you combine.
  • On a long one-variable multiple-choice question, graph the original once and test each choice for full-curve overlap.
  • Start by hand for short rewrites, answers you have to write yourself, more than one variable, or a variable in a denominator. One shared point never proves two expressions equivalent.

Related lessons

When the equivalent form you need is a product, use Factor algebraic expressions. When an identity or a given factor pins down unknown constants, use Use polynomial identities, factors, and unknown coefficients. For roots or fractional powers, go to Rewrite radicals and rational exponents, and when a variable shows up in a denominator, go to Rewrite rational expressions and preserve restrictions.

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