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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1951 : Algebra Ii

Divide 3x^3+5x^2-x+8 by x^2-1 .

Possible Answers:

$3x+5+\frac{2x+13}{x^2-1}$

$x^2-1-\frac{3x+5}{2x+13}$

$2x+13+\frac{3x+5}{x^2-1}$

$3x+5+\frac{x^2+1}{2x-13}$

$3x+5-\frac{2x+13}{x^2-1}$

Correct answer:

$3x+5+\frac{2x+13}{x^2-1}$

Explanation:

First, set up the division as the following:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}$

Look at the leading term x^2 in the divisor and 3x^3 in the dividend. Divide 3x^3 by x^2 gives ; therefore, put on the top:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}$

Then take that and multiply it by the divisor, x^2-1 , to get 3x^3-3x . Place that 3x^3-3x under the division sign:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \end{array}$

Subtract the dividend by that same 3x^3-3x and place the result at the bottom. The new result is 5x^2+2x+8 , which is the new dividend.

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}$

Now, 5x^2 is the new leading term of the dividend. Dividing 5x^2 by x^2 gives 5. Therefore, put 5 on top:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}$

Multiply that 5 by the divisor and place the result, 5x^2-5 , at the bottom:

Perform the usual subtraction:

Therefore the answer is 3x+5 with a remainder of 2x+13 , or $3x+5+\frac{2x+13}{x^2-1}$ .

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Example Question #12 : How To Divide Polynomials

Simplify the expression:

$\frac{x^{2}y^{15}z^{-2}}{x^{5}y^{4}}$

Possible Answers:

$\frac{y^{9}}{z^{2}}$

$\frac{y^{11}}{x^{3}z^{2}}$

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

$\frac{y^{12}}{x^{2}z^{2}}$

The fraction cannot be simplified further.

Correct answer:

$\frac{y^{11}}{x^{3}z^{2}}$

Explanation:

When dividing polynomials, subtract the exponent of the variable in the numberator by the exponent of the same variable in the denominator.

If the power is negative, move the variable to the denominator instead.

First move the negative power in the numerator to the denominator:

$\frac{x^{2}y^{15}}{x^{5}y^{4}z^{2}}$

Then subtract the powers of the like variables:

$\frac{y^{11}}{x^{3}z^{2}}$

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Example Question #1 : Simplifying Expressions

Simplify:

$x^{3}\times{x^{4}}+x$

Possible Answers:

$x^{8}$

$x^{13}$

$x^{12} + x$

$x^{7}+x$

$3x^{8}$

Correct answer:

$x^{7}+x$

Explanation:

$x^{m}\times{x^{n}} = x^{m+n}$ . However, $x^{7}+x$ cannot be simplified any further because the terms have different exponents.

(Like terms are terms that have the same variables with the same exponents. Only like terms can be combined together.)

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Example Question #21 : Simplifying Expressions

Simplify:

$\left (\frac{6x^{2}}{t} \right )^{-3}$

Possible Answers:

$\frac{216 t^{3} } { x^{6} }$

$\frac{18 t^{3} } {x^{6} }$

$\frac{t^{3} } { 216 x^{6} }$

$\frac{t^{3} } { 216 x^{5} }$

$\frac{t^{3} } {18 x^{6} }$

Correct answer:

$\frac{t^{3} } { 216 x^{6} }$

Explanation:

Apply the laws of exponents as follows:

$\left (\frac{6x^{2}}{t} \right )^{-3}$

$= \left (\frac{t} {6x^{2}}\right )^{3}$

$= \frac{t^{3} } { \left ( 6x^{2} \right ) ^{3} }$

$= \frac{t^{3} } { 6 ^{3} \left ( x^{2} \right ) ^{3}}$

$= \frac{t^{3} } { 216 \left ( x^{2 \cdot 3} \right ) }$

$= \frac{t^{3} } { 216 x^{6} }$

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Example Question #1952 : Algebra Ii

Simplify the expression:

$\frac{17}{x^{2}-x-6} + \frac{x+5}{x-3}$

Possible Answers:

$\frac{x^{2}+7x+27}{x+2}$

$\frac{x^{2}+7x+10}{x^{2}-x-6}$

$\frac{x^{2}+7x+27}{x-3}$

$\frac{x^{2}+7x+27}{x^{2}-x-6}$

$\frac{x+22}{x^{2}-x-6}$

Correct answer:

$\frac{x^{2}+7x+27}{x^{2}-x-6}$

Explanation:

$\frac{17}{x^{2}-x-6} + \frac{x+5}{x-3}$

1. Factor $x^{2}-x-6$

(x+2)(x-3)

***notice that the two fractions now share a factor in the denominator***

$\frac{17}{(x+2)(x-3)} + \frac{x+5}{x-3}$

2. Create a common denominator between the two terms

$\frac{17}{(x+2)(x-3)} + \frac{(x+5)(x+2)}{(x-3)(x+2)}$

3. Simplify

$=\frac{17}{x^{2}-x-6} + \frac{x^{2}+7x+10}{x^{2}-x-6}$

$=\frac{17 +x^{2}+7x+10}{x^{2}-x-6}$

$=\frac{x^{2}+7x+27}{x^{2}-x-6}$

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Example Question #62 : Expressions

Add and simplify the following rational expression:

$\frac{\sqrt{5}}{\sqrt{7}}+\frac{\sqrt{10}}{\sqrt{3}}$

Possible Answers:

$17\sqrt{\frac{5}{21}}$

$\frac{\sqrt{85}}{\sqrt{21}}$

No real solution

$\small \frac{\sqrt{35}}{7}+\frac{\sqrt{30}}{3}$

$\sqrt{\frac{3}{2}}$

Correct answer:

$\small \frac{\sqrt{35}}{7}+\frac{\sqrt{30}}{3}$

Explanation:

To add any fractions together, they must first have a common denominator. We can obtain a common denominator of $\sqrt{21}$ if we multiply the first fraction by $\frac{\sqrt{3}}{\sqrt{3}}$ and the second one by $\frac{\sqrt{7}}{\sqrt{7}}$ . We therefore obtain:

$\frac{\sqrt{5}}{\sqrt{7}}\times\frac{\sqrt{3}}{\sqrt{3}} + \frac{\sqrt{10}}{\sqrt{3}}\times\frac{\sqrt{7}}{\sqrt{7}}$

$\small \frac{\sqrt{15}}{\sqrt{21}}+\frac{\sqrt{70}}{\sqrt{21}}=\frac{\sqrt{15}+\sqrt{70}}{\sqrt{21}}$

From there, we need to take out the radical in the denominator by multiplying by $\small \frac{\sqrt{21}}{\sqrt{21}}$ , as follows:

$\small \small \frac{\sqrt{15}+\sqrt{70}}{\sqrt{21}}\times\frac{\sqrt{21}}{\sqrt{21}}=\frac{\sqrt{315}+\sqrt{1470}}{21}$

From here, we can simplify the radicals above by finding their prime factors:

$\small \sqrt{315}=\sqrt{3*3*5*7}=3\sqrt{35}$

and

$\small \sqrt{1470}=\sqrt{2*5*3*7*7}=7\sqrt{30}$ .

We are therefore left with $\small \frac{3\sqrt{35}+7\sqrt{30}}{21}$ , which can be separated and reduced to our final answer,

$\small \frac{\sqrt{35}}{7}+\frac{\sqrt{30}}{3}$

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Example Question #1953 : Algebra Ii

Simply: $(x^{4/3})^{1/2}$

Possible Answers:

$x^{2/3}$

$x^{11/6}$

$x^{-1}$

Correct answer:

$x^{2/3}$

Explanation:

In this form, the exponents are multiplied: $\left(\frac{4}{3}\right)\left(\frac{1}{2}\right)=\frac{4}{6}=\frac{2}{3}$ .

In multiplication problems, the exponents are added.

In division problems, the exponents are subtracted.

It is important to know the difference.

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Example Question #4662 : Algebra 1

Find the product: $5g(g^{2} - 4)$

Possible Answers:

$5g^{3}-20$

$-16g^{2}$

$5g^{2}-20g$

$5g^{3}-20g$

$5g^{2}-20$

Correct answer:

$5g^{3}-20g$

Explanation:

times $g^{2}$ gives us $5g^{3}$ , while times 4 gives us 20g . So it equals $5g^{3}-20g$ .

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Example Question #1954 : Algebra Ii

Distribute:

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

Possible Answers:

$8x^{3}+3x^{2}-4x$

$8x^{3}+12x^{2}-16x$

$8x^{3}+12x^{2}-16$

$8x^{2}+12-16$

$8x^{3}+12x^{2}+16x$

Correct answer:

$8x^{3}+12x^{2}-16x$

Explanation:

Be sure to distribute the along with its coefficient.

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Example Question #4262 : Algebra 1

Evaluate the following:

(2x^2+3x-4) + (3x^2-6x+7)

Possible Answers:

5x^2+3x+3

5x^2-3x-3

5x^2-3x+3

5x^2+3x-3

Correct answer:

5x^2-3x+3

Explanation:

With this problem, you need to take the trinomials out of parentheses and combine like terms. Since the two trinomials are being added together, you can remove the parentheses without needing to change any signs:

(2x^2+3x-4) + (3x^2-6x+7)

2x^2+3x-4 + 3x^2-6x+7

The next step is to combine like terms, based on the variables. You have two terms with x^2 , two terms with , and two terms with no variable. Make sure to pay attention to plus and minus signs with each term when combining like terms:

5x^2-3x+3

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1951 : Algebra Ii

Example Question #12 : How To Divide Polynomials

Example Question #1 : Simplifying Expressions

Example Question #21 : Simplifying Expressions

Example Question #1952 : Algebra Ii

Example Question #62 : Expressions

Example Question #1953 : Algebra Ii

Example Question #4662 : Algebra 1

Example Question #1954 : Algebra Ii

Example Question #4262 : Algebra 1

All Algebra II Resources

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