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Algebra II : Intermediate Single-Variable Algebra

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Example Questions

Example Question #21 : Simplifying Polynomials

Simplify the following expressions by combining like terms:

$-\left ( -3x^{3} +2x^{2} -xy -5x-13\right ) + x^{5} -4x^{4} -2x^{3} -9xy$

Possible Answers:

$x^{5}-4x^{4} +x^{3}-2x^{2} -13$

$x^{5}-4x^{4}-5x^{3}-2x^{2}-10xy+13$

$x^{5}-4x^{4} +x^{3} +2x^{2}-8xy -13$

$x^{5}-4x^{4}-5x^{3}-2x^{2} +13$

$x^{5}-4x^{4}+x^{3}-2x^{2}-8xy+5x+13$

Correct answer:

$x^{5}-4x^{4}+x^{3}-2x^{2}-8xy+5x+13$

Explanation:

Distribute the negative sign through all terms in the parentheses:

$3x^{3} - 2x^{2}+xy +5x +13$

Add the second half of the expression, $x^{5}-4x^{4}-2x^{3}-9xy$ to get:

$x^{5}-4x^{4}+x^{3}-2x^{2}-8xy+5x+13$

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Example Question #22 : Intermediate Single Variable Algebra

Write as a polynomial in standard form: $(3x - 2y)^{3}$

Possible Answers:

$27x^{3} - 18 x^{2} y + 12 xy^{2} - 8y^{3}$

$27x^{3} - 54 x^{2} y + 36 xy^{2} - 8y^{3}$

$27x^{3} - 8y^{3}$

$27x^{3} + 18 x^{2} y - 12 xy^{2} - 8y^{3}$

$27x^{3} + 54 x^{2} y - 36 xy^{2} - 8y^{3}$

Correct answer:

$27x^{3} - 54 x^{2} y + 36 xy^{2} - 8y^{3}$

Explanation:

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

Replace: A = 3x, B = 2y :

$(3x -2y)^{3}$

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

$= 3^{3}x^{3} - 3 \cdot 3^{2} \cdot 2 \cdot x^{2} y + 3 \cdot 3 \cdot 2^{2} \cdot xy^{2} - 2^{3}y^{3}$

$= 27x^{3} - 54 x^{2} y + 36 xy^{2} - 8y^{3}$

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Example Question #21 : Intermediate Single Variable Algebra

Combine:

3x^2y^3z+8m^4n^5-12x^2y^3z-2m^4n^5

Possible Answers:

11x^2y^3z

36x^2y^3z-16m^4n^5

21m^4n^5x^2y^3z

-9x^2y^3z+6m^4n^5

Correct answer:

-9x^2y^3z+6m^4n^5

Explanation:

When combining polynomials, only combine like terms. With the like terms, combine the coefficients. Your answer is -9x^2y^3z+6m^4n^5 .

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Example Question #24 : Intermediate Single Variable Algebra

Simplify this expression: (2y)(4x^2z^2 + 2a^2b^2) + (5x^2z^2y + 7a^2b^2y)

Possible Answers:

11y + a^2 b^2 y+13 x^2 y z^2

Not able to simplify further

11 a^2 b^2 y+13 x^2 y z^2

11 a^2 b^2 y + 15 x^2 y z^2

a^2 b^2 y+13 x^2 y z^2

Correct answer:

11 a^2 b^2 y+13 x^2 y z^2

Explanation:

Don't be scared by complex terms! First, we follow our order of operations and multiply the into the first binomial. Then, we check to see if the variables are alike. If they match perfectly, we can add and subtract their coefficients just like we could if the expression was 3x + 3x .

Remember, a variable is always a variable, no matter how complex! In this problem, the terms match after we follow our order of operations! So we just add the coefficients of the matching terms and we get our answer: 11 a^2 b^2 y+13 x^2 y z^2

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Example Question #21 : Polynomials

Simplify the following expression.

(6x^2-9x+1)-(4x^2+2x-4)

Possible Answers:

2x^2-11x+5

10x^2-7x-3

-2x^2+11x-5

x^2+8x-12

8x^2-7x+3

Correct answer:

2x^2-11x+5

Explanation:

(6x^2-9x+1)-(4x^2+2x-4)

First, we will need to distribute the minus sign.

6x^2-9x+1-4x^2-2x+4)

Then, combine like terms.

(6x^2-4x^2)+(-9x-2x)+(1+4)

2x^2-11x+5

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Example Question #21 : Intermediate Single Variable Algebra

Divide:

$(x^{3} -23x+10) \div (x-5)$

Possible Answers:

$x^{2} - 5x + 16 + \frac{87}{x-5}$

$x^{2}+1+\frac{12}{x-5}$

$x^{2} + 5x + 2$

$x^{2} + 5x + 2 + \frac{20}{x-5}$

$x^{2} - 5x + 16 - \frac{73}{x-5}$

Correct answer:

$x^{2} + 5x + 2 + \frac{20}{x-5}$

Explanation:

First, rewrite this problem so that the missing $x^{2}$ term is replaced by $0x^{2}$

$(x^{3} + 0x^{2} -23x+10) \div (x-5)$

Divide the leading coefficients:

$x^{3} \div x = x^{2}$ , the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

$x^{2} (x-5) = x^{3} -5x^{2}$

$(x^{3} + 0x^{2} -23x+10) - (x^{3} -5x^{2}) = 5x^{2}-23x+10$

Repeat this process with each difference:

$5x^{2} \div x = 5x$ , the second term of the quotient

$5x (x-5) = 5 x^{2} -25$

$5x^{2}-23x+10 - ( 5x^{2} -25x) = 2x+10$

One more time:

$2x \div x = 2$ , the third term of the quotient

2 (x-5) = 2x-10

2x+10 - (2x-10) = 20 , the remainder

The quotient is $x^{2} + 5x + 2$ and the remainder is ; this can be rewritten as a quotient of

$x^{2} + 5x + 2 + \frac{20}{x-5}$

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Example Question #21 : Intermediate Single Variable Algebra

Divide:

$\left (x^{2} + 6x -7 \right )\div (x+3)$

Possible Answers:

$x- 3 + \frac{2}{x+3}$

$x+ 3 + \frac{2}{x+3}$

$x- 3 - \frac{16}{x+3}$

$x+ 3 - \frac{16}{x+3}$

$x+ 3 + \frac{16}{x+3}$

Correct answer:

$x+ 3 - \frac{16}{x+3}$

Explanation:

Divide the leading coefficients to get the first term of the quotient:

$x^{2} \div x = x$ , the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

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

$\left (x^{2} + 6x -7 \right ) - (x^{2}+3x) = 3x -7$

Repeat these steps with the differences until the difference is an integer. As it turns out, we need to repeat only once:

$3x\div x = 3$ , the second term of the quotient

3 (x+3) = 3x + 9

(3x -7) - (3x + 9) = -16 , the remainder

Putting it all together, the quotient can be written as $x+ 3 - \frac{16}{x+3}$ .

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Example Question #22 : Intermediate Single Variable Algebra

$\textup{What is the product of }2x\textup{ and }3x^{2}-5x+2\textup{?}$

Possible Answers:

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

$6x^{2}-10x+4$

$6x^{3}+10x^{2}+4x$

$6x^{3}-10x^{2}+4x$

$3x^{2}-10x^{2}+2$

Correct answer:

$6x^{3}-10x^{2}+4x$

Explanation:

$\textup{Distribute: Multiply each term in the polynomial by }2x:$

$2x\left ( 3x^{2}-5x+2\right )=2x(3x^{2})+2x(-5x)+2x(2)$

$6x^{3}-10x^{2}+4x$

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Example Question #23 : Intermediate Single Variable Algebra

$4x^{2}y\left ( 2xy + 6xy^{2} + 3y^{{3}}\right ) =$

Possible Answers:

$8x^{3}y^{2} + 24x^{3}y^{3} + 12x^{4}y^{2}$

$8x^{3}y^{2} + 24x^{3}y^{3} + 12x^{2}y^{4}$

$32x^{3}y^{3} + 12x^{2}y^{4}$

$8x^{3}y^{2} + 24x^{2}y^{2} + 12x^{3}$

$8x^{2}y^{2} + 16x^{2}y^{3} + 24x^{2}y^{4}$

Correct answer:

$8x^{3}y^{2} + 24x^{3}y^{3} + 12x^{2}y^{4}$

Explanation:

Use the distributive property to obtain each term:

$4x^{2}y\left ( 2xy + 6xy^{2} + 3y^{{3}}\right ) =$

$4x^{2}y(2xy) + 4x^{2}y(6xy^{2}) + 4x^{2}y(3y^{3}) =$

$8x^{3}y^{2} + 24x^{3}y^{3} + 12x^{2}y^{4}$

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Example Question #24 : Intermediate Single Variable Algebra

Simplify the polynomials. Assume that no variables equal zero.

$\frac{12n^4p^3}{36n^6p^7}$

Possible Answers:

$\frac{6n^2}{18p^4}$

$\frac{n^2}{p^4}$

$\frac{1}{4n^2p^4}$

$\frac{4n^2}{p^4}$

Correct answer:

$\frac{1}{4n^2p^4}$

Explanation:

It is easiest to break this problem into groups, group the constant terms together, then group the N variables and group the P variables, like so.

$\frac{12}{36}\cdot \frac{n^4}{n^6}\cdot \frac{p^3}{p^7}$

Then reduce each fraction

$\frac{1}{4}\cdot \frac{1}{n^2}\cdot\frac{1}{p^4} = \frac{1}{4n^2p^4}$

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #21 : Simplifying Polynomials

Example Question #22 : Intermediate Single Variable Algebra

Example Question #21 : Intermediate Single Variable Algebra

Example Question #24 : Intermediate Single Variable Algebra

Example Question #21 : Polynomials

Example Question #21 : Intermediate Single Variable Algebra

Example Question #21 : Intermediate Single Variable Algebra

Example Question #22 : Intermediate Single Variable Algebra

Example Question #23 : Intermediate Single Variable Algebra

Example Question #24 : Intermediate Single Variable Algebra

All Algebra II Resources

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