GMAT Math : Simplifying Algebraic Expressions

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Example Question #31 : Simplifying Algebraic Expressions

Simplify: $\left (\sqrt[6]{x^{24}} \right )^{-4}$ . Assume is nonnegative.

Possible Answers:

$x^{26}$

$\frac{1}{x^{80}}$

$\frac{1}{x^{16}}$

$\frac{1}{x^{ 6}}$

Correct answer:

$\frac{1}{x^{16}}$

Explanation:

$\sqrt[m]{x^{n}}$ is equal to $x^{\frac{m}{n}}$ , $\left (x^{m} \right )^{n} = x^{mn}$ , and $x^{-n}=\frac{1}{x^{n}}$ ; combine these three properties as follows:

$\left (\sqrt[6]{x^{24}} \right )^{-4} = \left ( x^{\frac{24}{6}} \right )^{-4} = \left ( x^{4} \right )^{-4} = x^{4 \cdot (-4)} = x^{-16} = \frac{1}{x^{16}}$

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Example Question #32 : Simplifying Algebraic Expressions

Simplify:

$(x+7)(x^{2}-7x+49) + (x-7)(x^{2}+7x+49)$

Possible Answers:

$2x^{3} + 14x ^{2}+ 98x+ 686$

$2x^{3}$

$2x^{3} + 98x$

$2x^{3} - 14x ^{2}+ 98x- 686$

$2x^{3} + 686$

Correct answer:

$2x^{3}$

Explanation:

The problem is easier if you recognize the two products $(x+7)(x^{2}-7x+49)$ and $(x-7)(x^{2}+7x+49)$ as the factorizations of the sum and difference of two cubes, respectively:

$(x+7)(x^{2}-7x+49)= (x+7)(x^{2}-7 \cdot x +7^{2}) = x^{3}+7^{3}= x^{3}+343$

$(x-7)(x^{2}+7x+49)= (x-7)(x^{2}+7 \cdot x +7^{2}) = x^{3}-7^{3}= x^{3}-343$

Therefore,

$(x+7)(x^{2}-7x+49) + (x-7)(x^{2}+7x+49)$

$=\left ( x^{3}+343 \right )+ \left ( x^{3}-343 \right )$

$= x^{3}+x^{3}+343 -343$

$=2x^{3}$

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Example Question #31 : Simplifying Algebraic Expressions

Simplify:

$\frac{1- \frac{2}{x}}{1+\frac{4}{x^{2}}}$

Possible Answers:

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

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

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

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

$\frac{x}{x+2}$

Correct answer:

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

Explanation:

Multiply both numerator and denominator by common denominator $x^{2}$ :

$\frac{1- \frac{2}{x}}{1+\frac{4}{x^{2}}} = \frac{\left (1- \frac{2}{x} \right )x^{2}}{\left (1+\frac{4}{x^{2}} \right )x^{2}} =\frac{x^{2}-2x}{x^{2}+4}$

The denominator is a prime polynomial, so this is in simplest form. This is the correct choice.

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Example Question #261 : Algebra

Simplify: $\sqrt{x^{6}+6x^{4}+9x^{2}}$

You may assume all radicands assume positive values.

Possible Answers:

$x^{3}+3x$

$x \sqrt{x^{2}+3}$

$x \sqrt{x +3}$

$x^{3}+3\sqrt{x}$

$x^{2} \sqrt{x}+3 \sqrt{x}$

Correct answer:

$x^{3}+3x$

Explanation:

Factor the radicand first:

$x^{6}+6x^{4}+9x^{2}$

$=x^{2} \left (x^{4}+6x^{2}+9 \right )$

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

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

Therefore,

$\sqrt{x^{6}+6x^{4}+9x^{2}}$

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

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

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

$= x^{3}+3x$

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Example Question #262 : Algebra

Simplify: $\sqrt{x^{5}-10x^{4}+25x^{3}}$ .

You may assume all radicands assume positive values.

Possible Answers:

$x^{2}\sqrt{ x }- 10 x ^{2}+25x\sqrt{ x }$

$x^{2}\sqrt{ x }- 10 x\sqrt{ x }+25\sqrt{ x }$

$x^{2}\sqrt{ x }- 5 x\sqrt{ x }$

$x^{2}\sqrt{ x }- x ^{2}\sqrt{10}+5x\sqrt{ x }$

$x^{2} - 5 x$

Correct answer:

$x^{2}\sqrt{ x }- 5 x\sqrt{ x }$

Explanation:

Factor the radicand first:

$x^{5}-10x^{4}+25x^{3}$

$= x^{3} \left (x^{2}-10x +25 \right )$

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

Therefore,

$\sqrt{x^{5}-10x^{4}+25x^{3}}$

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

$=\sqrt{ x^{2} } \cdot \sqrt{ \left (x- 5 \right )^{2}}\cdot \sqrt{ x }$

$=x (x- 5 ) \sqrt{ x }$

$= (x^{2}- 5 x) \sqrt{ x }$

$= x^{2}\sqrt{ x }- 5 x\sqrt{ x }$

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Example Question #1341 : Problem Solving Questions

Which of the following is equal to the expression $\sqrt{abc} \cdot \sqrt{bcd}\cdot \sqrt{cde}$ ?

Assume all variables stand for nonnegative values.

Possible Answers:

None of the other responses is correct.

$bcd \sqrt{a c e}$

abcde

$ae \sqrt{bcd}$

$\sqrt{abcde}$

Correct answer:

$bcd \sqrt{a c e}$

Explanation:

$\sqrt{abc} \cdot \sqrt{bcd}\cdot \sqrt{cde}$

$= \sqrt{abc\cdot bcd \cdot cde}$

$= \sqrt{a\cdot b\cdot b\cdot c\cdot c\cdot c\cdot d\cdot d \cdot e}$

$= \sqrt{ b\cdot b } \cdot \sqrt{c \cdot c\cdot c} \cdot \sqrt{d \cdot d} \cdot \sqrt{a \cdot e}$

$=bcd \sqrt{a c e}$

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Example Question #263 : Algebra

Define an operation $\ast$ as follows:

For all real numbers a, b ,

$a \ast b = a^{2}+ b^{2}$

Which of the following is equal to $(y-x) \ast (x-y)$ ?

Possible Answers:

$2x^{2} - 4xy + 2 y^{2}$

-4xy

4xy

$2x^{2} + 2 y^{2}$

Correct answer:

$2x^{2} - 4xy + 2 y^{2}$

Explanation:

$a \ast b = a^{2}+ b^{2}$

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

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

$=x ^{2}+x ^{2}-2xy-2xy +y ^{2}+y ^{2}$

$=2x ^{2}-4xy +2y ^{2}$

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Example Question #33 : Simplifying Algebraic Expressions

Simplify the following expression:

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

Possible Answers:

$\small{x^2+2x+14}$

$\small{-x^2+2x+14}$

$\small{x^2+12x+14}$

$\small{x^2+2x+4}$

$\small{x^2+2x-14}$

Correct answer:

$\small{x^2+2x+14}$

Explanation:

Begin by dividing the fraction. We are dividing by x^2 , so simply subtract two from each of the exponents in the numerator.

So this:

$\small\frac{5x^2-3x^4+7x^3}{x^2}+4x^2-5x+9$

becomes this:

$\large$ $\small5-3x^2+7x+4x^2-5x+9$

Then, just combine like terms to finish up.

$\small{\color{Red} 5}{\color{Green} -3x^2}+{\color{Blue} 7x}+{\color{Green} 4x^2}{\color{Blue} -5x}+{\color{Red} 9}$

In red: 5 + 9 =14

In green: -3x^2 + 4x^2= x^2

In blue: 7x + (-5x) = 2x

So, we are left with:

$\small{x^2+2x+14}$

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Example Question #1342 : Problem Solving Questions

Simplify

$\frac{x^{2}+4x+3}{x^{2}-6x-7}$

Possible Answers:

$\frac{x+3}{x-7}$

$\frac{x-7}{x+3}$

$\frac{x+1}{x-7}$

$\frac{x-1}{x-7}$

$\frac{x+1}{x-3}$

Correct answer:

$\frac{x+3}{x-7}$

Explanation:

In order to simplify $\frac{x^{2}+4x+3}{x^{2}-6x-7}$ , factor both the numerator and denominator and then simplify:

$\frac{x^{2}+4x+3}{x^{2}-6x-7}$

Factoring the numerator and denominator we get two binomial functions on top and on bottom.

$=\frac{(x+1)(x+3)}{(x+1)(x-7)}$

Since (x+1) appears in both the numerator and the denominator they cancel eachother out and we are left with,

$=\frac{x+3}{x-7}$

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Example Question #265 : Algebra

Simplify

$\frac{x^{2}+7x+10}{x^{2}+2x-15}$

Possible Answers:

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

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

$\frac{x+5}{x-3}$

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

$\frac{x+2}{x+5}$

Correct answer:

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

Explanation:

In order to simplify $\frac{x^{2}+7x+10}{x^{2}+2x-15}$ , first factor the numerator and denominator and then cancel out factors as needed:

$\frac{x^{2}+7x+10}{x^{2}+2x-15}$

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

Since (x+5) appears in both the numerator and denominator, they cancel eachother out.

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

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GMAT Math : Simplifying Algebraic Expressions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #31 : Simplifying Algebraic Expressions

Example Question #32 : Simplifying Algebraic Expressions

Example Question #31 : Simplifying Algebraic Expressions

Example Question #261 : Algebra

Example Question #262 : Algebra

Example Question #1341 : Problem Solving Questions

Example Question #263 : Algebra

Example Question #33 : Simplifying Algebraic Expressions

Example Question #1342 : Problem Solving Questions

Example Question #265 : Algebra

All GMAT Math Resources

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