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High School Math : Intermediate Single-Variable Algebra

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Example Questions

Example Question #1 : Simplifying Polynomials

Simplify the following polynomial:

$\frac{-9x^{-2}y^3z^{-1}}{3x^{-4}yz^2}$

Possible Answers:

$\frac{-3x^{2}y^4}{z^3}$

$\frac{-9x^{4}y^2}{z^3}$

$\frac{-9x^{2}y^2}{z^3}$

$\frac{-3x^{2}y^2}{z^6}$

$\frac{-3x^{2}y^2}{z^3}$

Correct answer:

$\frac{-3x^{2}y^2}{z^3}$

Explanation:

To simplify the polynomial, begin by rearranging the terms to have positive exponents:

$\frac{-9x^{-2}y^3z^{-1}}{3x^{-4}yz^2}$

$\frac{-9x^{4}y^3}{3x^{2}yz^2z^1}$

Now, combine like terms:

$\frac{-9x^{2}y^2}{3z^3}$

Simplify the integers:

$\frac{-3x^{2}y^2}{z^3}$

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Example Question #11 : Simplifying Polynomials

Simplify the following polynomial:

$(\frac{2y^{-2}}{-2})^{-1}(\frac{m^2n}{y})^{-2}$

Possible Answers:

$\frac{-y^3}{m^4n^2}$

$\frac{y^3}{m^4n^2}$

$\frac{-y^4}{m^4n^2}$

$\frac{y^4}{m^4n^2}$

$\frac{y^2}{m^4n^2}$

Correct answer:

$\frac{-y^4}{m^4n^2}$

Explanation:

Begin by reversing the numerator and denominator so that the exponents are positive:

$(\frac{2y^{-2}}{-2})^{-1}(\frac{m^2n}{y})^{-2}$

$(\frac{-2}{2y^{-2}})^{1}(\frac{y}{m^2n})^{2}$

$(\frac{-2y^{2}}{2})^{1}(\frac{y}{m^2n})^{2}$

Square the right side of the expression and multiply:

$(\frac{-2y^{2}}{2})(\frac{y^2}{m^4n^2})$

$\frac{-2y^{4}}{2m^4n^2}$

Simplify:

$\frac{-y^{4}}{m^4n^2}$

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Example Question #261 : Algebra Ii

Simplify the following polynomial:

$\frac{6n^{2x+3}}{3n^{x-4}}$

Possible Answers:

$n^{x+7}$

$3n^{x+7}$

$2n^{x+7}$

$2n^{x+9}$

$n^{x+9}$

Correct answer:

$2n^{x+7}$

Explanation:

Begin by simplifying the integers:

$\frac{6n^{2x+3}}{3n^{x-4}}$

$\frac{2n^{2x+3}}{n^{x-4}}$

Subtract the exponent in the denominator from the exponent in the numerator:

(2x+3)-(x-4)=x+7

$2n^{x+7}$

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Example Question #262 : Algebra Ii

Simplify the following polynomial:

$x^{-1}y^2(x^{-3}y-xy^{-3}+x^2y^{-4})$

Possible Answers:

$\frac{y^3}{x^3}-\frac{1}{y^2}+\frac{x}{y^2}$

$\frac{y^2}{x^4}-\frac{2}{y}+\frac{x}{y^2}$

$\frac{y^2}{x^4}-\frac{1}{y}+\frac{x}{y^2}$

$\frac{y^3}{x^4}-\frac{3}{y}+\frac{x^3}{y^2}$

$\frac{y^3}{x^4}-\frac{1}{y}+\frac{x}{y^2}$

Correct answer:

$\frac{y^3}{x^4}-\frac{1}{y}+\frac{x}{y^2}$

Explanation:

Begin by multiplying the terms:

$x^{-1}y^2(x^{-3}y-xy^{-3}+x^2y^{-4})$

$x^{-4}y^3-y^{-1}+xy^{-2}$

Convert into fraction form:

$\frac{y^3}{x^4}-\frac{1}{y}+\frac{x}{y^2}$

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Example Question #263 : Algebra Ii

Simplify the following polynomial:

$(4x^{a+2}-3y^{3a})^2$

Possible Answers:

$16x^{2a+4}-24x^{a+2}y^{3a}+9y^{6a}$

$16x^{a+4}-24x^{a+2}y^{3a}+9y^{6a}$

$16x^{a+4}-24x^{a+2}y^{3a}+9y^{3a}$

$16x^{a+4}-24x^{2a+2}y^{3a}+9y^{6a}$

$16x^{a+4}-24x^{3a+2}y^{2a}+9y^{6a}$

Correct answer:

$16x^{2a+4}-24x^{a+2}y^{3a}+9y^{6a}$

Explanation:

Use the FOIL method to multiply the terms: F (first) O (outer) I (inner) L (last)

$(4x^{a+2}-3y^{3a})^2$

$(4x^{a+2}-3y^{3a})(4x^{a+2}-3y^{3a})$

$16x^{2a+4}-24x^{a+2}3y^{3a}+9y^{6a}$

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Example Question #11 : Simplifying Polynomials

Simplify the following polynomial:

$(x^{n+1}+y^{2n-1})(x^{n+1}-y^{2n-1})$

Possible Answers:

$x^{2n+2}-y^{4n-4}$

$x^{n+2}-y^{4n-2}$

$x^{2n+2}-y^{2n-2}$

$x^{2n+2}-y^{4n-2}$

$x^{2n+1}-y^{4n-2}$

Correct answer:

$x^{2n+2}-y^{4n-2}$

Explanation:

Use the FOIL method to multiply the terms: F (first) O (outer) I (inner) L (last)

$(x^{n+1}+y^{2n-1})(x^{n+1}-y^{2n-1})$

$(x^{2n+2}+x^{n+1}y^{2n-1}-x^{n+1}y^{2n-1}-y^{4n-2})$

Combine like terms:

$x^{2n+2}-y^{4n-2}$

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Example Question #261 : Algebra Ii

If f(x)=x-3 and $g(x)=x^{2}-x$ , what is g(f(x)) ?

Possible Answers:

$x^{2}-7x+12$

$x^{2}-3$

$x^{2}-x-3$

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

$x^{3}-4x^{2}-3x$

Correct answer:

$x^{2}-7x+12$

Explanation:

g(f(x)) is a composite function solved by substituting f(x) into g(x) :

$(x-3)^{2}-(x-3)=x^{2}-6x+9-x+3=x^{2}-7x+12$

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Example Question #1 : Factoring Polynomials

Factor $x^{3}-64$

Possible Answers:

Cannot be Factored

(x-4)(x^2+4x+16)

(x-4)(x^2-4x-16)

(x+4)(x^2-4x+16)

Correct answer:

(x-4)(x^2+4x+16)

Explanation:

Use the difference of perfect cubes equation:

a^3-b^3=(a-b)(a^2+ab+b^2)

In $x^{3}-64$ ,

a=x and $b=\sqrt[3]{64}=4$

(x-4)(x^2+(x)(4)+4^2)

(x-4)(x^2+4x+16)

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Example Question #2 : Factoring Polynomials

Factor the polynomial completely and solve for .

Possible Answers:

4x (8x^2 - 2) = 0

$x = \left \{0, \frac{1}{2} \right \}$

8x(4x^2 -1)

$x = \left \{ -\frac{1}{2}, 0, \frac{1}{2} \right \}$

8x(2x+1)(2x-1)

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

8(4x^3- x)

x = 0

32x(x+1)(x-1)

$x = \left \{ -1, 0,1\right \}$

Correct answer:

8x(2x+1)(2x-1)

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

Explanation:

To factor and solve for in the equation

Factor out of the equation

$\rightarrow 8x(4x^2-1)$

Use the "difference of squares" technique to factor the parenthetical term, which provides the completely factored equation:

$\rightarrow 8x(2x+1)(2x-1)$

Any value that causes any one of the three terms , 2x+1 , and 2x -1 to be will be a solution to the equation, therefore

$x = \left \{-\frac{1}{2}, 0, \frac{1}{2} \right \}$

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Example Question #3 : Factoring Polynomials

Factor the following expression:

x^2+xy+x=

Possible Answers:

x^2+x(y+1)

x(x+y)

x+y+1

x(x+y+1)

x+y

Correct answer:

x(x+y+1)

Explanation:

You can see that each term in the equation has an "x", therefore by factoring "x" from each term you can get that the equation equals x(x+y+1) .

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High School Math : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Simplifying Polynomials

Example Question #11 : Simplifying Polynomials

Example Question #261 : Algebra Ii

Example Question #262 : Algebra Ii

Example Question #263 : Algebra Ii

Example Question #11 : Simplifying Polynomials

Example Question #261 : Algebra Ii

Example Question #1 : Factoring Polynomials

Example Question #2 : Factoring Polynomials

Example Question #3 : Factoring Polynomials

All High School Math Resources

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