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Algebra II : Simplifying Expressions

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Example Questions

Example Question #1932 : Algebra Ii

Simplify x(4 – x) – x(3 – x).

Possible Answers:

x²

Correct answer:

Explanation:

You must multiply out the first set of parenthesis (distribute) and you get 4x – x². Then multiply out the second set and you get –3x + x². Combine like terms and you get x.

x(4 – x) – x(3 – x)

4x – x² – x(3 – x)

4x – x² – (3x – x²)

4x – x² – 3x + x² = x

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Example Question #1 : How To Divide Trinomials

Divide:

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

Possible Answers:

x-1

x+2

Correct answer:

x-1

Explanation:

Factor the numerator and denominator:

$\frac{3x^2+3x-6}{3x+6}=\frac{3(x+2)(x-1)}{3(x+2)}$

Cancel the factors that appear in both the numerator and the denominator:

$\frac{3(x+2)(x-1)}{3(x+2)}=\frac{x-1}{1}=x-1$

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Example Question #1 : Simplifying Expressions

Simplify: $\frac{5p^{7}}{25p^{^{4}}}$

Possible Answers:

$\frac{p^{7/4}}{5}$

$\frac{p}{5}$

$\frac{p^{3}}{5}$

$\frac{p^{-3}}{5}$

$5p^{3}$

Correct answer:

$\frac{p^{3}}{5}$

Explanation:

$p^{7}$ and $p^{4}$ cancel out, leaving $p^{3}$ in the numerator. 5 and 25 cancel out, leaving 5 in the denominator

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Example Question #1 : Simplifying Expressions

Simplify the following:

$\frac{4x+6x^2}{2x}$

Possible Answers:

2x(2-3x)

2+3x

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

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

$\frac{1}{x}-1\frac{1}{2}$

Correct answer:

2+3x

Explanation:

$\frac{4x+6x^2}{2x}$

First, let us factor the numerator:

4x+6x^2=x(4+6x)=2x(2+3x)

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

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Example Question #3 : Simplifying Expressions

Find the product:

$\small 7n(8n-2)$

Possible Answers:

$\small n^2-14n$

$\small 56n^2-14n$

$\small n-14$

$\small n^2+14$

Correct answer:

$\small 56n^2-14n$

Explanation:

First, mulitply the mononomial by the first term of the polynomial:

$\small 7n\times8n\ = 56n^2$

Second, multiply the monomial by the second term of the polynomial:

$\small 7n\times (-2)\ = -14n$

Add the terms together:

$\small 56n^2\ +\ (-14n)\ = 56n^2-14n$

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Example Question #1 : Simplifying Expressions

Multiply, expressing the product in simplest form:

$\small \frac{15x^{5}}{14y^{3}} \cdot \frac{21y^{2} }{25x^{3}}$

Possible Answers:

$\small \small \small \small \frac{10 x^{2}}{9y }$

$\small \small \small \small \frac{9 y}{10x^{2} }$

$\small \small \small \small \small \frac{10 y}{9x^{2} }$

$\small \small \frac{9 x^{2}}{10 y}$

$\small \small \small \frac{9 x^{2}y}{10 }$

Correct answer:

$\small \small \frac{9 x^{2}}{10 y}$

Explanation:

Cross-cancel the coefficients by dividing both 15 and 25 by 5, and both 14 and 21 by 7:

$\small \small \small \small \frac{15x^{5}}{14y^{3}} \cdot \frac{21y^{2} }{25x^{3}} = \small \small \small \small \frac{3x^{5}}{2y^{3}} \cdot \frac{3y^{2} }{5x^{3}} = \frac{3x^{5} \cdot 3y^{2}}{5x^{3} \cdot 2y^{3}} = \frac{3 \cdot 3 \cdot x^{5} y^{2}}{5\cdot 2\cdot x^{3} y^{3}} = \frac{9 x^{5} y^{2}}{10 x^{3} y^{3}}$

Now use the quotient rule on the variables by subtracting exponents:

$\small \frac{9 x^{5} y^{2}}{10 x^{3} y^{3}} = \frac{9 x^{5-3}}{10 y^{3-2}} = \frac{9 x^{2}}{10 y^{1}} = \frac{9 x^{2}}{10 y}$

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Example Question #4772 : Algebra 1

Simplify the following:

$\frac{2xy^2}{3yz^2} \times \frac{15y^2z}{14x^2y}$

Possible Answers:

$\frac{5xy^2}{7z}$

$\frac{5xz}{7y^2}$

$\frac{5y^2}{7xz}$

$\frac{7y^2}{5x^2z}$

Correct answer:

$\frac{5y^2}{7xz}$

Explanation:

In this problem, you have two fractions being multiplied. You can first simplify the coefficients in the numerators and denominators. You can divide and cancel the 2 and 14 each by 2, and the 3 and 15 each by 3:

$\frac{2xy^2}{3yz^2} \times \frac{15y^2z}{14x^2y} = \frac{xy^2}{yz^2} \times \frac{5y^2z}{7x^2y}$

You can multiply the two numerators and two denominators, keeping in mind that when multiplying like variables with exponents, you simplify by adding the exponents together:

$\frac{xy^2}{yz^2} \times \frac{5y^2z}{7x^2y} = \frac{5xy^4z}{7x^2y^2z^2}$

Any variables that are both in the numerator and denominator can be simplified by subtracting the numerator's exponent by the denominator's exponent. If you end up with a negative exponent in the numerator, you can move the variable to the denominator to keep the exponent positive:

$\frac{5xy^4z}{7x^2y^2z^2} = \frac{5y^2}{7xz}$

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Example Question #5 : Simplifying Expressions

Factor the expression:

$\small \small 2xy^3-14xy^2+9xy$

Possible Answers:

$\small 2xy(y^2-7xy+9)$

$\small xy^3(2-14xy^2+9xy)$

$\small x(2y^3-14y^2+9y)$

$\small 2xy^3(-14xy^2+9xy)$

$\small xy(2y^2-14y+9)$

Correct answer:

$\small xy(2y^2-14y+9)$

Explanation:

To find the greatest common factor, we need to break each term into its prime factors:

$\small 2xy^3=2 \cdot x \cdot y \cdot y \cdot y$

$\small 14xy^2 = 2 \cdot 7 \cdot x \cdot y \cdot y$

$\small 9xy = 3 \cdot 3 \cdot x \cdot y$

Looking at which terms all three expressions have in common; thus, the GCF is $\small xy$ . We then factor this out: $\small \small xy(2y^2 -14y +9)$ .

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Example Question #2 : Simplifying Expressions

Expand: 8x(3x+7)

Possible Answers:

24x^2 + 56x

24x^2 + 56

24x + 56

11x^2 + 15x

11x + 15

Correct answer:

24x^2 + 56x

Explanation:

To expand, multiply 8x by both terms in the expression (3x + 7).

8x multiplied by 3x is 24x².

8x multiplied by 7 is 56x.

Therefore, 8x(3x + 7) = 24x² + 56x.

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Example Question #1 : Binomials

Simplify:

-5(x+15)-3x

Possible Answers:

2x-75

-8x+15

2x+15

None of the other answers are correct.

-8x-75

Correct answer:

-8x-75

Explanation:

First, distribute –5 through the parentheses by multiplying both terms by –5.

-5(x+15)-3x=-5x-75-3x

Then, combine the like-termed variables (–5x and –3x).

-5x-75-3x=-8x-75

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Algebra II : Simplifying Expressions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1932 : Algebra Ii

Example Question #1 : How To Divide Trinomials

Example Question #1 : Simplifying Expressions

Example Question #1 : Simplifying Expressions

Example Question #3 : Simplifying Expressions

Example Question #1 : Simplifying Expressions

Example Question #4772 : Algebra 1

Example Question #5 : Simplifying Expressions

Example Question #2 : Simplifying Expressions

Example Question #1 : Binomials

All Algebra II Resources

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