Distribute two linear factors
Practice problem
Which expression is equivalent to
Why this matters on the SAT
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
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.
Two expressions are equivalent when they give the same value for every allowed value of their variables. For example,
The two sides look different, but distributing the turns the left side into the right side, so they give the same output for every .
You'll recognize this kind of question from wording like:
None of these asks you to find a value of . You're changing the form, or picking the matching form, and 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.
Start with full-curve overlap when the question looks like the opening example:
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, and both equal when , but at they give and . So a curve that only crosses the original proves nothing. You need the whole curve to match.
Start by hand in these cases instead:
Related: For more practice comparing expressions by graph, see Equivalent expressions by graph overlap.
Which method would you start with for each of these, and why? First, simplify . Second, pick which of four long choices matches a one-variable expression built from several products.
The distributive property says that a factor outside parentheses multiplies every term inside:
It works the same way with subtraction:
A minus sign in front of parentheses works like a factor of . Writing the in makes it clear that every sign inside flips:
Once the parentheses are gone, combine like terms, terms with exactly the same variable part. and are like terms. and aren't, because the powers of are different.
Here are both steps together:
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.
Rewrite . Why does the -term come out positive?
Turning into combines unlike terms. One has and the other has , so their variable parts don’t match, and is already as simple as it gets. Plugging in shows the problem: the original is , but is .
A polynomial is a sum of terms whose exponents are whole numbers, like . Adding, subtracting, and multiplying polynomials uses the same two moves over and over: distribute completely, then combine like terms.
When you add, the parentheses can come straight off. Then combine the matching terms:
Subtraction is where signs go missing. The minus sign applies to the whole second polynomial, so every one of its terms flips:
Look at the last term. Subtracting gives .
When you multiply two polynomials, every term in one has to multiply every term in the other. Two binomials give four products:
Write out all four products before you combine anything. That's what keeps the middle term, here , from going missing. The same every-term rule works for bigger polynomials too.
Rewrite . Keep the minus sign in view while you take off the second set of parentheses, then group the like terms.
What did you get for the Try it yourself problem, and which signs changed?
Worked example
Which expression is equivalent to
Step 1
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:
and
So the answer is A, and you didn't have to do the long expansion to find it.
Step 2
Now let's confirm it by hand. The algebra also shows where a wrong choice like B comes from. Multiply each term in by each term in :
Step 3
A square is a product of two copies, so it has a middle term: means , not . Put both expanded products in, and keep the square in parentheses until you've handled the minus sign in front of it:
Step 4
Group the -terms, the -terms, and the constants:
The algebra agrees with the graph. The answer is A.
Choice B comes from writing as , 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 , so all three signs flip: .
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.
Practice problem
Which expression is equivalent to
Practice problem
Which expression is equivalent to
Practice problem
The expression
is equivalent to , where , , and are constants. What is the value of ?
Finish the lesson
Finish the remaining questions correctly to complete this lesson.
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.
Next lesson
Combine powers exactly by matching bases and applying each exponent rule to the complete expression.
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