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SAT Math : SAT Mathematics

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Example Questions

Example Question #2 : Polynomials

Given a♦b = (a+b)/(a-b) and b♦a = (b+a)/(b-a), which of the following statement(s) is(are) true:

I. a♦b = -(b♦a)

II. (a♦b)(b♦a) = (a♦b)²

III. a♦b + b♦a = 0

Possible Answers:

II & III

I and III

I and II

I, II and III

I only

Correct answer:

I and III

Explanation:

Notice that - (a-b) = b-a, so statement I & III are true after substituting the expression. Substitute the expression for statement II gives ((a+b)/(a-b))((a+b)/(b-a))=((a+b)(b+a))/((-1)(a-b)(a-b))=-1 〖(a+b)〗²/〖(a-b)〗² =-((a+b)/(a-b))² = -(a♦b)² ≠ (a♦b)²

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Example Question #2151 : Sat Mathematics

If a positive integer a is divided by 7, the remainder is 4. What is the remainder if 3a + 5 is divided by 3?

Possible Answers:

Correct answer:

Explanation:

The best way to solve this problem is to plug in an appropriate value for a. For example, plug-in 11 for a because 11 divided by 7 will give us a remainder of 4.

Then 3a + 5, where a = 11, gives us 38. Then 38 divided by 3 gives a remainder of 2.

The algebra method is as follows:

a divided by 7 gives us some positive integer b, with a remainder of 4.

Thus,

a / 7 = b 4/7

a / 7 = (7b + 4) / 7

a = (7b + 4)

then 3a + 5 = 3 (7b + 4) + 5

(3a+5)/3 = [3(7b + 4) + 5] / 3

= (7b + 4) + 5/3

The first half of this expression (7b + 4) is a positive integer, but the second half of this expression (5/3) gives us a remainder of 2.

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Example Question #3 : How To Divide Polynomials

Polydivision1

Possible Answers:

100

Correct answer:

Explanation:

Polydivision2

Polydivision4

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Example Question #3 : Multiplying And Dividing Polynomials

Simplify: $\frac{6x^7y^3z^9}{3x^6y^3z}$

Possible Answers:

xyz^8

6xyz^8

xyz

xz^8

2xz^8

Correct answer:

2xz^8

Explanation:

Cancel by subtracting the exponents of like terms:

$\frac{6x^7y^3z^9}{3x^6y^3z} = 2x^{7-6}y^{3-3}z^{9-1}=2xy^0z^8=2xz^8$

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Example Question #12 : Polynomials

Divide $x^{3} + 125y ^{3}$ by x+ 5y .

Possible Answers:

$x ^{2}+ 5xy + 25y^{2}$

$x ^{2}+10xy + 25y^{2}$

$x ^{2} + 25y^{2}$

$x ^{2}-10xy + 25y^{2}$

$x ^{2}-5xy + 25y^{2}$

Correct answer:

$x ^{2}+ 5xy + 25y^{2}$

Explanation:

It is not necessary to work a long division if you recognize $x^{3} + 125y ^{3}$ as the sum of two perfect cube expressions:

$x^{3} + 125y ^{3} = x^{3} + 5^{3} y ^{3} = x^{3} + (5y )^{3}$

A sum of cubes can be factored according to the pattern

$A^{3}+ B^{3} = (A+B) (A^{2}+ AB +B^{2})$ ,

so, setting A = x, B = 5y ,

$x^{3} + (5y )^{3} =(x+5y) [x ^{2}+ x \cdot 5y + (5y)^{2}]$

$=(x+5y)(x ^{2}+ 5xy + 25y^{2})$

Therefore,

$\frac{x^{3} + 125y ^{3}}{x+ 5y} = \frac{(x+5y)(x ^{2}+ 5xy + 25y^{2})}{x+ 5y} = x ^{2}+ 5xy + 25y^{2}$

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Example Question #374 : Algebra

By what expression can 3x+7 be multiplied to yield the product $6x^{3} +20x^{2} +5x-21$ ?

Possible Answers:

$2x^{2 }+ x - 3$

$2x^{2 } - 3$

$2x^{2 }-x - 3$

$2x^{2 }+ 2x - 3$

Correct answer:

$2x^{2 }+ 2x - 3$

Explanation:

Divide $6x^{3} +20x^{2} +5x-21$ by 3x+7 by setting up a long division.

Divide the lead term of the dividend, $6x^{3}$ , by that of the divisor, ; the result is $\frac{6x^{3}}{3x} = 2 x^{2}$

Enter that as the first term of the quotient. Multiply this by the divisor:

$2 x^{2} (3x+7) = 2 x^{2} (3x)+2 x^{2} ( 7) = 2x^{3} + 14x^{2}$

Subtract this from the dividend. This is shown in the figure below.

Division poly

Repeat the process with the new difference:

$\frac{6x^{2}}{3x} = 2 x$

$2 x (3x+7) = 2 x (3x)+2 x ( 7) = 2x^{2} + 14x$

Division poly

Repeating:

$\frac{-9x}{3x} = -3$

-3 (3x+7) = -3 (3x)+ (-3) ( 7) = -9x+ 21

Division poly

The quotient - and the correct response - is $2x^{2 }+ 2x - 3$ .

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Example Question #1 : How To Multiply Polynomials

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

and

$G(x) = x^{2} + 5$

What is FG(x) ?

Possible Answers:

$(FG)(x) = x^{5} + x^{4} - x - 2$

$(FG)(x) = x^{3} - x - 3$

$(FG)(x) = x^{3} + 2x^{2} - x + 7$

$(FG)(x) = x^{5} + x^{4} +4x^{3} + 7x^{2} - 5x +10$

$(FG)(x) = x^{5} + x^{4} - x^{3} + 2x^{2} - 5x -10$

Correct answer:

$(FG)(x) = x^{5} + x^{4} +4x^{3} + 7x^{2} - 5x +10$

Explanation:

$(FG)(x) = F(x)G(x)$ so we multiply the two function to get the answer. We use $x^{m}x^{n} = x^{m+n}$

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Example Question #2152 : Sat Mathematics

Find the product:

( x^4-5x+7)(x^4+2x+1)

Possible Answers:

x^4+8x^2+10x-9

x^6+5x^5+3x^4+x^2

x^8-3x^5+8x^4-10x^2+9x+7

x^8+7x^4-10x^2

x^7+9x^4-10x^2+8

Correct answer:

x^8-3x^5+8x^4-10x^2+9x+7

Explanation:

Find the product:

( x^4-5x+7)(x^4+2x+1)

Step 1: Use the distributive property.

x^4(x^4+2x+1)-5x(x^4+2x+1)+7(x^4+2x+1)

x^8+2x^5+x^4-5x^5-10x^2-5x+7x^4+14x+7

Step 2: Combine like terms.

x^8+2x^5-5x^5+x^4+7x^4-10x^2-5x+14x+7

x^8-3x^5+8x^4-10x^2+9x+7

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Example Question #12 : Polynomials

represents a positive quantity; represents a negative quantity.

$x^{4} = 116$

$y^{2} = 11$

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

Possible Answers:

The correct answer is not among the other choices.

237

Correct answer:

Explanation:

The first two binomials are the difference and the sum of the same two expressions, which, when multiplied, yield the difference of their squares:

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

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

Again, a sum is multiplied by a difference to yield a difference of squares, which by the Power of a Power Property, is equal to:

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

$= x^{2\cdot 2} - y^{2\cdot 2}$

$= x^{4} - y^{4}$

$y^{2} = 11$ , so by the Power of a Power Property,

$(y^{2} )^{2}= 11^{2}$

$y^{2\cdot 2} = 121$

$y^{4} = 121$

Also, $x^{4} = 116$ , so we can now substitute accordingly:

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

$= x^{4} - y^{4}$

= 116 - 121

= -5

Note that the signs of and are actually irrelevant to the problem.

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Example Question #14 : Polynomials

represents a positive quantity; represents a negative quantity.

$x^{6} = 625$

$y ^{6} = 64$

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

Possible Answers:

Correct answer:

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, by way of the Power of a Power Property,

$(x^{3})^{ 2 } = x^{3 \cdot 2} = x^{6 }$ , making $x^{3}$ a square root of $x^{6}$ , or 625; since is positive, so is $x^{3}$ , so

$x^{3} = \sqrt{625} = 25$ .

Similarly, $y ^{3}$ is a square root of $y ^{6}$ , or 64; since is negative, so is $y^{3}$ (as an odd power of a negative number is negative), so

$y^{3} = - \sqrt{64} = -8$ .

Therefore, substituting:

$(x-y) (x^{2}+xy+y^{2}) = x^{3} - y^{3} = 25 - (-8) = 25+8= 33$ .

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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 : Polynomials

Example Question #2151 : Sat Mathematics

Example Question #3 : How To Divide Polynomials

Example Question #3 : Multiplying And Dividing Polynomials

Example Question #12 : Polynomials

Example Question #374 : Algebra

Example Question #1 : How To Multiply Polynomials

Example Question #2152 : Sat Mathematics

Example Question #12 : Polynomials

Example Question #14 : Polynomials

All SAT Math Resources

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