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Algebra II : Algebra II

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Example Questions

Example Question #4382 : Algebra 1

Combine:

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

Possible Answers:

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

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

11x^2y^3z

36x^2y^3z-16m^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 #2 : Dividing Polynomials

Divide:

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

Possible Answers:

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

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

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

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

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

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 : Simplifying Polynomials

Divide:

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

Possible Answers:

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

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

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

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

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

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 #23 : Simplifying Polynomials

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

Possible Answers:

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

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

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

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

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

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 #21 : Intermediate Single Variable Algebra

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

Possible Answers:

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

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

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

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

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

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 #1162 : Algebra Ii

Simplify the polynomials. Assume that no variables equal zero.

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

Possible Answers:

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

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

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

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

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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Example Question #1163 : Algebra Ii

Simplify the polynomial. Assume that no variable equals zero.

(2x-2y)^2

Possible Answers:

4x^2-8xy+4y^2

4x^2+4y^2

4x^2+8xy-4y^2

4x^2-4y^2

Correct answer:

4x^2-8xy+4y^2

Explanation:

It is important to remember that ;

(a-b)^2 = (a-b)(a-b)

$\\ (2x-2y)^2=(2x-2y)(2x-2y) \\ =4x^2-4xy-4xy+4y^2 \\ =4x^2-8xy+4y^2$

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Example Question #1164 : Algebra Ii

Simplify: 16x^4-2x^3-8x^2+10x

Possible Answers:

2x(8x^3-x^2-4x+5x)

x(8x^3-x^2-4x+5)

2(8x^3-x^2-4x+5)

2x(8x^3-x^2-4x+5)

Correct answer:

2x(8x^3-x^2-4x+5)

Explanation:

To simplify this expression, first identify what all the terms have in common. In this case, it's 2x. Now, factor that out from each of the terms to get your answer: 2x(8x^3-x^2-4x+5) .

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4382 : Algebra 1

Example Question #24 : Intermediate Single Variable Algebra

Example Question #21 : Polynomials

Example Question #2 : Dividing Polynomials

Example Question #21 : Simplifying Polynomials

Example Question #23 : Simplifying Polynomials

Example Question #21 : Intermediate Single Variable Algebra

Example Question #1162 : Algebra Ii

Example Question #1163 : Algebra Ii

Example Question #1164 : Algebra Ii

All Algebra II Resources

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