Variables and Exponents - ISEE Upper Level Mathematics Achievement

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Question

Rewrite the polynomial in standard form:

$9x^{4} - 3x ^{2} + 8x^{5} -2x + 7 - 4x^{3}$

Answer

The degree of a term of a polynomial with one variable is the exponent of that variable. The terms of a polynomial in standard form are written in descending order of degree. Therefore, we rearrange the terms by their exponent, from 5 down to 0, noting that we can rewrite the and constant terms with exponents 1 and 0, respectively:

$9x^{4} - 3x ^{2} + 8x^{5} -2x + 7 - 4x^{3}$

$= 9x^{4} - 3x ^{2} + 8x^{5} -2x ^{1} + 7x ^{0} - 4x^{3}$

$=8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x ^{1}+ 7x^{0}$

$=8x^{5} + 9x^{4}- 4x^{3} - 3x ^{2} -2x+ 7$

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Question

Simplify:

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

Answer

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

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

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

$=\frac{4x^{2}y^{2}}{x^{4}y^{4}}$

$=\frac{4}{x^{2}y^{2}}$

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Question

Assume that . Simplify:

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

Answer

$\frac{(x^{2}+y^{2})^{2} + (x^{2} + y^{2})^{2}}{x^{4}y^{4}}$

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

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

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

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Question

If , simplify:

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

Answer

$\frac{(x-y)^2+4xy}{x+y}=\frac{(x^2-2xy+y^2)+4xy}{x+y}$

$=\frac{(x^2-2xy+4xy+y^2)}{x+y}$

$=\frac{(x^2+2xy+y^2)}{x+y}$

$=\frac{(x+y)^2}{x+y}=x+y$

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Question

If , simplify:

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

Answer

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

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

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

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

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

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

$=\frac{2(x^4+3)}{2}$

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Question

Define

What is $g(x ^{2} + 3x)$ ?

Answer

Substitute $x ^{2} + 3x$ for in the definition:

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

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

$g(x ^{2} + 3x ) = 2 x ^{2} + 6x - 5$

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Question

Add:

Answer

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Question

Simplify:

Answer

Start by reordering the expression to group like-terms together.

Combine like-terms to simplify.

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Question

Simplify:

Answer

We can expand the first term using FOIL:

Reorder the expression to group like-terms together.

Simplify by combining like-terms.

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Question

Simplify:

Answer

Expand each term by using FOIL:

Rearrange to group like-terms together.

Simplify by combining like-terms.

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Question

Simplify the following:

$\frac{p^{6}}{p^{3}}$

Answer

To divide a variable with an exponent, you can write out all the multiplies, which for this question would look like:

$\frac{pppppp}{ppp}$

Then cancel out the matching p's, leaving:

$ppp=p^{3}$

Or, if the top and bottom are the same variable, you can simply subtract the bottom exponent from the top exponent.

$p^{6-3}=p^{3}$

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Question

Simplify the following:

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

Answer

To divide variables with exponents, either:

Write out the multiplies, then reduce

$\frac{yy}{yyyyy}=\frac{1}{y^{3}}$

or subtract the bottom exponent from the top, like so

$y^{2-5}=y^{-3}=\frac{1}{y^{3}}$

Remember, negative exponent means dividing. $y^{-3}$ would also be correct, but you should know that is equivalent to $\frac{1}{y^{3}}$

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Question

If , simplify:

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

Answer

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

$=\frac{3x(x^2+1)\left [ (x+1)(x-1) \right ]}{x(x^2-1)}$

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

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

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Question

Simplify:

$\frac{x^6y^2z^3}{x^3y^4z^3}$

Answer

$\frac{x^6y^2z^3}{x^3y^4z^3}=x^{6-3}y^{2-4}z^{3-3}=x^3y^{-2}z^0=\frac{x^3}{y^2}$

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Question

Simplify if $xy eq \pm 2$ .

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

Answer

Start by factoring the numerator. can be removed from each term.

$\frac{x^4y^4-4x^2y^2}{(xy+2)(xy-2)}=\frac{x^2y^2(x^2y^2-4)}{(xy+2)(xy-2)}$

Next, expand the denominator.

$\frac{x^2y^2(x^2y^2-4)}{(xy+2)(xy-2)}=\frac{x^2y^2(x^2y^2-4)}{x^2y^2-4}$

Simplify by canceling terms.

$\frac{x^2y^2(x^2y^2-4)}{x^2y^2-4}=x^2y^2$

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Question

Simplify:

$\frac{2x^{-7}}{(3x)^{-2}}$

Answer

In the denominator, apply the exponent to the terms in parentheses.

$\frac{2x^{-7}}{(3x)^{-2}}=\frac{2x^{-7}}{(3^{-2}) (x^{-2})}$

Terms with negative exponents can be inverted within the fraction to eliminate the negative.

$\frac{2x^{-7}}{(3^{-2}) (x^{-2})}=\frac{2(3^2)(x^2)}{x^7}$

Simplify by canceling variables and mulitplying.

$\frac{2(3^2)(x^2)}{x^7}=\frac{18x^2}{x^7}=\frac{18}{x^5}$

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Question

Simplify:

$\frac{8x^{-2a}y^{2a}}{4x^ay^{-2a}}$

Answer

Break up the fraction into components with like-terms.

$\frac{8x^{-2a}y^{2a}}{4x^ay^{-2a}}=\frac{8}{4}\times \frac{x^{-2a}}{x^a}\times \frac{y^{2a}}{y^{-2a}}$

Simplify by combining exponents. Remember that exponents in the denominator will become negative.

$2\times x^{(-2a-a)}\times y^{(2a+2a)}$

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

Finally, put any negative exponent terms into the denominator.

$\frac{2y^{4a}}{x^{3a}}$

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Question

Simplify:

$\frac{(2xy)^{6}}{2x^{2} y^{8}}$

Answer

$\frac{(2xy)^{6}}{2x^{2} y^{8}}$

$= \frac{2^{6} \cdot x^{6}y^{6}}{2x^{2} y^{8}}$

$= \frac{64 x^{6-2}}{2 y^{8-6}}$

$= \frac{32 x^{4} }{ y^{2}}$

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Question

Divide the following:

$x^8 \div x^2$

Answer

To divide variables with exponents, we will use the following formula:

$x^a \div x^b = x^{a-b}$

Now, let’s combine the following:

$x^8 \cdot x^2 = x^{8-2}$

$x^8 \cdot x^2 = x^{6}$

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Question

Divide the following:

$\frac{a^2 b^4}{a b^2}$

Answer

When dividing variables, we can only divide like variable.

Also, when dividing variables with exponents, we subtract the exponents. In other words:

$\frac{a^x}{a^y} = a^{x-y}$

So, we get

$\frac{a^2 b^4}{a b^2}$

$\frac{a^2b^4}{a^1b^2}$

$a^{2-1}b^{4-2}$

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