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ISEE Upper Level Math : How to multiply exponential variables

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Example Question #1 :How To Multiply Exponential Variables

Simplify the following:

$(z^{3} x^{2} y)(x^{3} y^{4})$

Possible Answers:

$(zxy)^{13}$

$z^{3}x^{6}y^{4}$

$z^{3}x^{5}y^{5}$

$z^{3}xy^{3}$

None of the other answers

Correct answer:

$z^{3}x^{5}y^{5}$

Explanation:

To multiply variables with exponents, add the exponents. With multiple variables, simply add the exponents for each different variable.

Simplified:

$(z^{3})(x^{2+3})(y^{1+4})$

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Example Question #2 :How To Multiply Exponential Variables

Simplify the following:

$(3py^{2})(4p^{4})$

Possible Answers:

$12p^{5}y^{2}$

$12(py)^{8}$

$7p^{5}y^{2}$

$7py^{8}$

None of the other answers

Correct answer:

$12p^{5}y^{2}$

Explanation:

To multiply variables with exponents, add the exponents. When there are constants mixed in, multiply the constants separately and put back in the final result:

$(3py^{2})(4p^{4})$

$3\times4\times p^{1+4}\times y^{2}$

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Example Question #1 :How To Multiply Exponential Variables

Simplify the following:

$z^{2}\times z^{5}$

Possible Answers:

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

$z^{10}$

$z^{7}$

$z^{3}$

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

Correct answer:

$z^{7}$

Explanation:

To multiply variables with exponents, add the exponents. So,

$z^{2}\times z^{5}=z^{7}$

A longer way would be to write out all the multiplies the exponent tells us to do. This is a little clearer on why adding the exponents works but takes longer and isn't necessary once you understand the process.

$z^{2}\times z^{5}= zzzzzzz= z^{7}$

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Example Question #4 :How To Multiply Exponential Variables

Factor completely:

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

Possible Answers:

$\left(x ^{2}- 2 \right ) \left (x + 2 \right )$

$\left(x ^{2}- 2 \right ) \left (x - 2 \right )$

$x\left (x - 4 \right )\left (x - 1 \right )$

$x\left (x - 2 \right )\left (x - 2 \right )$

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

Correct answer:

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

Explanation:

is the greatest common factor of each term, so distribute it out:

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

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

$= x \left (x^{2}- 3x + 4 \right )$

We try to factor $x^{2}- 3x + 4$ by finding two integers with product 4 and sum. However, both of our possible factor pairs fail, since -2 + (-2) = -4 and -1+ (-4) = -5 .

$x \left (x^{2}- 3x + 4 \right )$ is the complete factorization.

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Example Question #5 :How To Multiply Exponential Variables

Multiply:

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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

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Example Question #6 :How To Multiply Exponential Variables

Factor completely:

$4 y^{2} - 100 y$

Possible Answers:

$4y ^{2}(y - 25)$

4y (y - 25)

$4 (y - 5)^{2}$

2 (y + 10) (4y - 5)

4 (y + 5) (y - 5)

Correct answer:

4y (y - 25)

Explanation:

The greatest common factor of the terms in $4 y^{2} - 100 y$ is, so factor that out:

$4 y^{2} - 100 y$

$= 4y\left ( \frac{4 y^{2}}{4y} - \frac{100 y}{4y} \right )$

$= 4y\left (y - 25 \right )$

Since all factors here are linear, this is the complete factorization.

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Example Question #7 :How To Multiply Exponential Variables

Exponentiate:

$\left (x + 0.5y \right) ^{3}$

Possible Answers:

$x^{3} + 0.125 y ^{3}$

$x^{3} + x^{2} y + 0.5 \ x y ^{2} + 0.125 y ^{3}$

$x^{3} + 1.5 x^{2} y + 0.75 \ x y ^{2} + 0.125 y ^{3}$

$x^{3} + 1.5 x^{2} y + 0.75 \ x y ^{2} + 1.25 y ^{3}$

$x^{3} + 0.5 x^{2} y + 0.25 \ x y ^{2} + 0.125 y ^{3}$

Correct answer:

$x^{3} + 1.5 x^{2} y + 0.75 \ x y ^{2} + 0.125 y ^{3}$

Explanation:

The cube of a sum pattern can be applied here:

$\left (x + 0.5y \right) ^{3}$

$= x^{3} + 3 \cdot x^{2}\cdot 0.5y + 3 \cdot x \cdot\left (0.5y \right) ^{2} + \left (0.5y \right) ^{3}$

$= x^{3} + 3\cdot 0.5 \cdot x^{2} \cdot y + 3 \cdot 0.5^{2} \cdot x \cdot y ^{2} + 0.5^{3} \cdot y ^{3}$

$= x^{3} + 3 \cdot 0.5 x^{2} y + 3 \cdot 0.25 \ x y ^{2} + 0.125 y ^{3}$

$= x^{3} + 1.5 x^{2} y + 0.75 \ x y ^{2} + 0.125 y ^{3}$

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Example Question #8 :How To Multiply Exponential Variables

Write in expanded form:

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

Possible Answers:

$8x^{3} + 6 x^{2} + 3x + \frac{1}{8}$

$8x^{3} + \frac{1}{8}$

$8x^{3} + 2 x^{2} + \frac{1}{2} x + \frac{1}{8}$

$8x^{3} + 6 x^{2} + \frac{3}{2} x + \frac{1}{8}$

$8x^{3} + x^{2} +x + \frac{1}{8}$

Correct answer:

$8x^{3} + 6 x^{2} + \frac{3}{2} x + \frac{1}{8}$

Explanation:

The cube of a sum pattern can be applied here:

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

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

$= 2^{3}\cdot x^{3} + 3\cdot 2^{2}\cdot \frac{1}{2} \cdot x^{2} + 3\cdot2 \cdot \left (\frac{1}{2} \right ) ^{2} \cdot x \left + \left ( \frac{1}{2} \right ) ^{3}$

$= 8x^{3} + 6 x^{2} + \frac{3}{2} x + \frac{1}{8}$

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Example Question #9 :How To Multiply Exponential Variables

Factor completely:

$x ^{2} - 11x + 20$

Possible Answers:

The expression cannot be factored.

(x - 4)(x - 5)

(x + 2)(x - 10)

(x - 4)(x + 5)

(x - 2)(x - 10)

Correct answer:

The expression cannot be factored.

Explanation:

A trinomial with leading coefficient $x^{2}$ can be factored by looking for two integers to fill in the boxes:

$(x + \square )(x + \square )$ .

The numbers should have product 20 and sum. However, all of the possible factor pairs fail:

-1 + (-20) = -21

-2 + (-10) = -12

-4 + (-5) = -9

The polynomial is prime.

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Example Question #10 :How To Multiply Exponential Variables

Factor completely:

$4x^{2} - 36x+ 81$

Possible Answers:

(2x-3)(2x-27)

(x-3)(4x-27)

$(2x-9)^{2}$

(4x-9)(x-9)

(4x-3)(x-27)

Correct answer:

$(2x-9)^{2}$

Explanation:

$4x^{2} - 36x+ 81$ can be seen to be a perfect square trinomial by taking the square root of the first and last terms, multiplying their product by 2, then comparing it to the second term:

$4x^{2} = \left ( 2x\right )^{2}$

$81 = 9^{2}$

$2 \cdot2x\cdot9= 36x$

Therefore,

$4x^{2} - 36x+ 81$

$=\left ( 2x \right ) ^{2} -2 \cdot 2 x \cdot9+ 9 ^{2}$

$=(2x-9)^{2}$

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ISEE Upper Level Math : How to multiply exponential variables

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #1 :How To Multiply Exponential Variables

Example Question #2 :How To Multiply Exponential Variables

Example Question #1 :How To Multiply Exponential Variables

Example Question #4 :How To Multiply Exponential Variables

Example Question #5 :How To Multiply Exponential Variables

Example Question #6 :How To Multiply Exponential Variables

Example Question #7 :How To Multiply Exponential Variables

Example Question #8 :How To Multiply Exponential Variables

Example Question #9 :How To Multiply Exponential Variables

Example Question #10 :How To Multiply Exponential Variables

All ISEE Upper Level Math Resources

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