Variables and Exponents

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ISEE Upper Level Quantitative Reasoning › Variables and Exponents

Questions 1 - 10

Simplify if $xy\neq \pm 2$ .

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

$x^{-1}y^2$

Explanation

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.

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

Simplify if $xy\neq \pm 2$ .

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

$x^{-1}y^2$

Explanation

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.

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

Multiply:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$x^{2} - 6xy +9y ^{2} - 1$

$x^{2} +9y ^{2} - 1$

$x^{2} + 6xy -9y ^{2} - 1$

$x^{2} + 3xy -9y ^{2} - 1$

$x^{2} - 3xy +9y ^{2} - 1$

Explanation

This can be achieved by using the pattern of difference of squares:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$=\left [ \left ( x -3y \right ) + 1\right ] \left \left[( x - 3y \right ) - 1\right ]$

$= ( x -3y ) ^{2} - 1 ^{2}$

$= (x -3y) ^{2} - 1$

Applying the binomial square pattern:

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

$= x^{2} - 6xy +9y ^{2} - 1$

Multiply:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$x^{2} - 6xy +9y ^{2} - 1$

$x^{2} +9y ^{2} - 1$

$x^{2} + 6xy -9y ^{2} - 1$

$x^{2} + 3xy -9y ^{2} - 1$

$x^{2} - 3xy +9y ^{2} - 1$

Explanation

This can be achieved by using the pattern of difference of squares:

$\left ( x -3y + 1\right ) \left ( x - 3y - 1\right )$

$=\left [ \left ( x -3y \right ) + 1\right ] \left \left[( x - 3y \right ) - 1\right ]$

$= ( x -3y ) ^{2} - 1 ^{2}$

$= (x -3y) ^{2} - 1$

Applying the binomial square pattern:

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

$= x^{2} - 6xy +9y ^{2} - 1$

Simplify the following:

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

$\frac{1}{y^{3}}$

$y^{3}$

$y^{0.4}$

$\frac{1}{3y}$

Explanation

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 $\frac{1}{y^{3}}$

Simplify the following:

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

$\frac{1}{y^{3}}$

$y^{3}$

$y^{0.4}$

$\frac{1}{3y}$

Explanation

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 $\frac{1}{y^{3}}$

Simplify:

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

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

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

$\frac{4}{xy}$

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

Explanation

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

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

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

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

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

Rewrite the polynomial in standard form:

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

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

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

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

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

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

Explanation

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: