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SAT Math : SAT Mathematics

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Example Questions

Example Question #2 : How To Multiply Polynomials

and represent positive quantities.

$x^{2} = 20$

$y^{2} = 5$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

Possible Answers:

$5 \sqrt{5}$

$45 \sqrt{15 }$

$35 \sqrt{15 }$

$35 \sqrt{5}$

$45\sqrt{5}$

Correct answer:

$35 \sqrt{5}$

Explanation:

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

$(x-y) (x^{2}+xy+y^{2}) = x^{3} - y^{3}$

Additionally,

$x^{2} = 20$ and is positive, so

$x = \sqrt{20}$

Using the product of radicals property, we see that

$x = \sqrt{4 \cdot 5} = \sqrt{4 } \cdot \sqrt{ 5} = 2 \sqrt{ 5}$

and

$x ^{3 } = x^{2} \cdot x = 20 \cdot 2 \sqrt{5} = 40 \sqrt{5}$

$y^{2} = 5$ and is positive, so

$y = \sqrt{5}$ ,

and

$y^{3}= y^{2} \cdot y = 5 \sqrt{5}$

Substituting for $x ^{3 }$ and $y^{3}$ , then collecting the like radicals,

$(x-y) (x^{2}+xy+y^{2})$

$= x^{3} - y^{3}$

$= 40 \sqrt{5}-5 \sqrt{5}$

$=( 40 -5 )\sqrt{5}$

$=35 \sqrt{5}$ .

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Example Question #1 : How To Subtract Polynomials

Simplify the following expression:

$(2q^{2}+4q+7)-(3q^{2}-5)$

Possible Answers:

$-3q^{2}+12$

$-q^{2}+4q+12$

-q^2+4q+2

$-5q^{2}+4q+2$

Correct answer:

$-q^{2}+4q+12$

Explanation:

This is not a FOIL problem, as we are adding rather than multiplying the terms in parentheses.

Add like terms together:

$2q^{2}-3q^{2}=-1q^{2}=-q^2$

has no like terms.

7-(-5)=12

Combine these terms into one expression to find the answer:

$-q^{2}+4q+12$

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Example Question #1 : How To Find The Solution To A Binomial Problem

Define an operation $\heartsuit$ on the set of real numbers as follows:

For all real m, n ,

$m \heartsuit n = \left ( m-4n\right )^{3}$

How else could this operation be defined?

Possible Answers:

$m \heartsuit n = m ^{3} +4 m ^{2} n - 16mn ^{2}- 64n ^{3}$

$m \heartsuit n = m ^{3} -4 m ^{2} n + 16mn ^{2}- 64n ^{3}$

$m \heartsuit n = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$

$m \heartsuit n = m ^{3} +12 m ^{2} n - 48mn ^{2}-64n ^{3}$

$m \heartsuit n = m ^{3} -64n ^{3}$

Correct answer:

$m \heartsuit n = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$

Explanation:

$\left ( m-4n\right )^{3}$ , as the cube of a binomial, can be rewritten using the following pattern:

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot m ^{2} \cdot 4n + 3 \cdot m \cdot (4n) ^{2}- (4n) ^{3}$

Applying the rules of exponents to simplify this:

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot 4 \cdot m ^{2} \cdot n + 3 \cdot m \cdot 4 ^{2}\cdot n ^{2}- 4^{3} \cdot n ^{3}$

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot 4 \cdot m ^{2} \cdot n + 3 \cdot 16 \cdot m\cdot n ^{2}- 4^{3} \cdot n ^{3}$

$\left ( m-4n\right )^{3} = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$

Therefore, the correct choice is that, alternatively stated,

$m \heartsuit n = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$ .

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Example Question #1 : Binomials

Solve for .

$\frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}$

Possible Answers:

Correct answer:

Explanation:

$\frac{x^{2}+2x-8}{x^{2}+5x-14}=\frac{3}{4}$

Factor the expression

numerator: find two numbers that add to 2 and multiply to -8 [use 4,-2]

denominator: find two numbers that add to 5 and multiply to -14 [use 7,-2]

new expression:

$\frac{(x+4)(x-2)}{(x+7)(x-2)}=\frac{3}{4}$

Cancel the x-2 and cross multiply.

4(x+4)=3(x+7)

4x+16=3x+21

x=5

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Example Question #1 : Binomials

If 〖(x+y)〗²= 144 and 〖(x-y)〗²= 64, what is the value of xy?

Possible Answers:

Correct answer:

Explanation:

We first expand each binomial to get x² + 2xy + y² = 144 and x² - 2xy + y² = 64. We then subtract the second equation from the first to find 4xy = 80. Finally, we divide each side by 4 to find xy = 20.

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Example Question #22 : Polynomials

Solve each problem and decide which is the best of the choices given.

What are the zeros of the following trinomial?

3x^3-30x^2+63x

Possible Answers:

x=7,3

x=0,7,3

x=0,-7,-3

x=-7,-3

x=3,7,-3

Correct answer:

x=0,7,3

Explanation:

First factor out a . Then the factors of the remaining polynomial,

3x(x^2-10x+21)=0 , are (x-7) and (x-3) .

Set everything equal to zero and you get x=3 , x=7 , and x=0 because you cant forget to set equal to zero.

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Example Question #1 : Trinomials

Subtract 2x^2-3x+6 from 4x^2+10x-11 .

Possible Answers:

-2x^2-13x+17

2x^2+13x-17

2x^2-13x+17

-2x^2-13x-17

Correct answer:

2x^2+13x-17

Explanation:

Step 1: We need to read the question carefully. It says subtract from. When you see the word "from", you read the question right to left.

I am subtracting the left equation from the right equation.

Step 2: We need to write the equation on the right minus the equation of the left.

4x^2+10x-11-(2x^2-3x+6)

Step 3: Distribute the minus sign in front of the parentheses:

4x^2+10x-11-2x^2+3x-6

Step 4: Combine like terms:

4x^2-2x^2=2x^2
10x+3x=13x
-11-6=-17

Step 5: Put all the terms together, starting with highest degree. The degree of the terms is the exponent. Here, the highest degree is 2 and lowest is zero.

The final equation is 2x^2+13x-17

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Example Question #24 : Polynomials

What is a possible value for x in x² – 12x + 36 = 0 ?

Possible Answers:

There is not enough information

–6

Correct answer:

Explanation:

You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative

(x –__)(x –__).

You should realize that 6 fits into both blanks.

You must now set each set of parenthesis equal to 0.

x – 6 = 0; x – 6 = 0

Solve both equations: x = 6

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Example Question #3 : Algebra

If r and t are constants and x² +rx +6=(x+2)(x+t), what is the value of r?

Possible Answers:

It cannot be determined from the given information.

Correct answer:

Explanation:

We first expand the right hand side as x²+2x+tx+2t and factor out the x terms to get x²+(2+t)x+2t. Next we set this equal to the original left hand side to get x²+rx +6=x²+(2+t)x+2t, and then we subtract x² from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.

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Example Question #4 : Factoring Polynomials

Solve for $x$ :

$2x^2-4=3 +5$

Possible Answers:

$\sqrt{6}$

$6$

$\pm \sqrt{12}$

$\pm \sqrt{6}$

$12$

Correct answer:

$\pm \sqrt{6}$

Explanation:

$2x^2-4=3 +5$

First, add 4 to both sides:

2x^2=12

Divide both sides by 2:

x^2=6

$x=\pm \sqrt{6}$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2 : How To Multiply Polynomials

Example Question #1 : How To Subtract Polynomials

Example Question #1 : How To Find The Solution To A Binomial Problem

Example Question #1 : Binomials

Example Question #1 : Binomials

Example Question #22 : Polynomials

Example Question #1 : Trinomials

Example Question #24 : Polynomials

Example Question #3 : Algebra

Example Question #4 : Factoring Polynomials

All SAT Math Resources

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