GMAT Math : Simplifying Algebraic Expressions

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Example Question #21 : Simplifying Algebraic Expressions

Which of the following is equal to the expresion $\sqrt{16x} - \sqrt{25x} + \sqrt{98x}$ ?

Possible Answers:

$6 \sqrt{2x}$

$12 \sqrt{x}$

$13 \sqrt{x}$

$7\sqrt{2x}-\sqrt{x}$

$7x\sqrt{2}-\sqrt{x}$

Correct answer:

$7\sqrt{2x}-\sqrt{x}$

Explanation:

$\sqrt{16x} - \sqrt{25x} + \sqrt{98x}$

$=\sqrt{16} \cdot \sqrt{x} - \sqrt{25} \cdot \sqrt{x} + \sqrt{49} \cdot \sqrt{2x}$

$=4\sqrt{x} - 5 \sqrt{x} +7 \sqrt{2x}$

$=\left (4 - 5 \right ) \sqrt{x} +7 \sqrt{2x}$

$=- \sqrt{x} +7 \sqrt{2x}$

$= 7 \sqrt{2x} - \sqrt{x}$

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Example Question #21 : Simplifying Algebraic Expressions

Simplify: $7x^{5} + 7x^{4}- 4 x - 4$

Possible Answers:

$14x^{9} - 8 x$

The polynomial cannot be simplified further.

$7x^{9} - 4 x - 4$

$14x^{9} - 4 x - 4$

$7x^{9} - 8 x$

Correct answer:

The polynomial cannot be simplified further.

Explanation:

None of the terms of the polynomial have the same degree - the exponents of in the terms are, in order, 5, 4, 1, and 0. Therefore, the four terms are all unlike terms, and none can be combined. The polynomial is already in simplified form.

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Example Question #23 : Simplifying Algebraic Expressions

Simplify:

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

Possible Answers:

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

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

$x^{3}-2 x^{2} + 2x- 1$

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

$x^{3}+4x^{2} -4x- 1$

Correct answer:

$x^{3}-2 x^{2} + 2x- 1$

Explanation:

$\left (x-1 \right )^{2}$

$= x^{2}- 2 \cdot 1 \cdot x + 1^{2}$

$= x^{2}- 2 x + 1$

$\left (x-1 \right )^{3}$

$= x^{3} - 3 \cdot 1 \cdot x^{2} + 3 \cdot 1^{2} \cdot x - 1^{3}$

$= x^{3} -3 x^{2} + 3 x - 1$

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

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

$= x^{3}+x^{2} -3 x^{2} + x -2x+3x- 1+1-1$

$= x^{3}-2 x^{2} + 2x- 1$

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Example Question #24 : Simplifying Algebraic Expressions

Simplify: $(x-7)^{2} - (x+14)^{2}$

Possible Answers:

14x-147

-42x -147

-21x -147

14x+245

-42x +245

Correct answer:

-42x -147

Explanation:

Use the perfect square trinomial pattern to expand, then collect like terms:

$(x-7)^{2} - (x+14)^{2}$

$=(x^{2} - 2\cdot 7 \cdot x +7^{2} ) - (x^{2} + 2\cdot 14 \cdot x +14^{2} )$

$=(x^{2} -14 x +49) - (x^{2} + 28x +196 )$

$=x^{2} -14 x +49 - x^{2} - 28x -196$

$=x^{2}- x^{2} -14 x- 28x +49 -196$

= -42x -147

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Example Question #25 : Simplifying Algebraic Expressions

Simplify: $(x+3)+(x+3)^{2}+ (x+3)^{3}$

Possible Answers:

$x^{3}+ 10x^{2} +34 x + 39$

$x^{3}+ 9 x^{2} + 27 x + 38$

$x^{3}+ 9x^{2} +34 x + 39$

$x^{3}+ 9 x^{2} + 27 x + 39$

$x^{3}+10 x^{2} + 27 x + 38$

Correct answer:

$x^{3}+ 10x^{2} +34 x + 39$

Explanation:

Expand $(x+3)^{2}$ and $(x+3)^{3}$ :

$(x+3)^{2}$

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

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

$(x+3)^{3}$

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

$= x^{3}+ 9 x^{2} + 27 x + 27$

Add:

$x^{3}+ 9 x^{2} + 27 x + 27$

$x^{2}+\; 6x +\; \; \; 9$

$\underline{ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x\; \; \; +3}$

$x^{3}+ 10x^{2} +34 x + 39$

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Example Question #1337 : Problem Solving Questions

Which of the following is equal to $\sqrt{6x-6} \cdot \sqrt{15x-15} \cdot \sqrt{35x-35}$ ?

Possible Answers:

$15x\sqrt{14x-14}-15\sqrt{14x-14}$

$14x\sqrt{15x-15}-14\sqrt{15x-15}$

$210 x\sqrt{x-1}-210 \sqrt{x-1}$

210x-210

$210x^{2}-420x+210$

Correct answer:

$15x\sqrt{14x-14}-15\sqrt{14x-14}$

Explanation:

Factor all radicands, multiply, then simplify:

$\sqrt{6x-6} \cdot \sqrt{15x-15} \cdot \sqrt{35x-35}$

$= \sqrt{2 \cdot 3\cdot (x-1)} \cdot \sqrt{3 \cdot 5 \cdot (x-1)} \cdot \sqrt{5\cdot 7\cdot (x-1)}$

$= \sqrt{2 \cdot 3\cdot3 \cdot 5 \cdot 5\cdot 7\cdot(x-1)\cdot (x-1)\cdot (x-1)}$

$=\sqrt{3\cdot3} \cdot \sqrt{5 \cdot 5} \cdot \sqrt{(x-1)\cdot (x-1)} \cdot \sqrt{2 \cdot 7 \cdot (x-1)}$

$=3 \cdot5 \cdot (x-1) \cdot \sqrt{14x-14}$

$= (15x-15) \cdot \sqrt{14x-14}$

$= 15x\sqrt{14x-14}-15\sqrt{14x-14}$

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Example Question #21 : Simplifying Algebraic Expressions

Simplify the expression:

(x-1)(x+1)+(x-2)(x+2)+(x-3)(x+3)

Possible Answers:

$3x^{2}-11$

$3x^{2}-6x -1 4$

$3x^{2}-12x -1 4$

$3x^{2}$

$3x^{2}-1 4$

Correct answer:

$3x^{2}-1 4$

Explanation:

Each of the three products added is a product of the sum and the difference of the same two expressions, so each can be simplified using that pattern:

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

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

$(x-3)(x+3)= x^{2}- 3^{2} = x^{2}-9$

Add:

(x-1)(x+1)+(x-2)(x+2)+(x-3)(x+3)

$=(x^{2}-1) +(x^{2}-4) +(x^{2}-9)$

$=x^{2}+x^{2}+x^{2}-1 -4 -9$

$=3x^{2}-1 4$

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Example Question #21 : Simplifying Algebraic Expressions

Simplify: $\sqrt[3]{ \sqrt[4]{x ^{20}} }$

You may assume is positive.

Possible Answers:

$x^{8}$

$x^{2} \sqrt[7]{x^{6}}$

$x^{13}$

None of the other choices gives the correct answer.

$x\sqrt[3]{x^{2}}$

Correct answer:

$x\sqrt[3]{x^{2}}$

Explanation:

$\sqrt[m]{x^{n}}$ is equal to $x^{\frac{m}{n}}$ , so $\sqrt[4]{x ^{20}}$ can be rewritten as $x^{\frac{20}{4}}$ or $x^{5}$ .

The original expression can be rewritten as $\sqrt[3]{ x ^{5}}$ , which itself can be rewritten by noting that 5 divided by 3 yields quotient 1 and remainder 2. Therefore, the simplified form is $x^{1} \sqrt[3]{x^{2}}$ , or $x \sqrt[3]{x^{2}}$ .

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Example Question #1340 : Problem Solving Questions

Simplify the expression:

$\sqrt[5]{x^{-8}y^{7}}$

Possible Answers:

$\frac{y \sqrt[5]{ x^{2} y^{2}} }{ x^{2}}$

$\frac{y \sqrt[5]{ x y^{2}} }{ x^{2}}$

$\frac{y \sqrt[5]{ x^{2} y^{2}} }{ x }$

$\frac{y \sqrt[5]{ x^{3} y^{2}} }{ x^{2}}$

$\frac{y \sqrt[5]{ x y^{2}} }{ x }$

Correct answer:

$\frac{y \sqrt[5]{ x^{2} y^{2}} }{ x^{2}}$

Explanation:

First, rewrite the radicand with positive exponents, then take the fifth root of the numerator and the denominator:

$\sqrt[5]{x^{-8}y^{7}} = \sqrt[5]{\frac{y^{7}}{x^{8}}} = \frac{ \sqrt[5]{ y^{7}} }{ \sqrt[5]{ x^{8}}}$

To simplify each expression, divide the exponent by the index 5, and note its quotient and remainder. 7 divided by 5 is 1 with remainder 2; 8 divided by 5 is 1 with remainder 3. Subsequently,

$\frac{ \sqrt[5]{ y^{7}} }{ \sqrt[5]{ x^{8}}} =\frac{y \sqrt[5]{ y^{2}} }{x \sqrt[5]{ x^{3}}}$

The denominator can be rationalized by multiplying both halves by $\sqrt[5]{ x^{2}}$ :

$\frac{y \sqrt[5]{ y^{2}}\cdot \sqrt[5]{ x^{2}} }{x \sqrt[5]{ x^{3}}\cdot \sqrt[5]{ x^{2}}}$

$=\frac{y \sqrt[5]{ x^{2} y^{2}} }{x \cdot \sqrt[5]{ x^{5}}}$

$=\frac{y \sqrt[5]{ x^{2} y^{2}} }{x \cdot x}$

$=\frac{y \sqrt[5]{ x^{2} y^{2}} }{ x^{2}}$

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Example Question #26 : Simplifying Algebraic Expressions

Simplify:

$(x+5)(x^{2}-5x+25) - (x-5)(x^{2}+5x+25)$

Possible Answers:

The correct answer is not among the other choices.

$25x^{2}+250$

$2x^{3} +10x$

$2x^{3} +50x$

$5x^{2}+250$

Correct answer:

The correct answer is not among the other choices.

Explanation:

The problem is easier if you recognize the two products $(x+5)(x^{2}-5x+25)$ and $(x-5)(x^{2}+5x+25)$ as the factorizations of the sum and difference of two cubes, respectively:

$(x+5)(x^{2}-5x+25) = (x+5)(x^{2}-5 \cdot x +5^{2}) = x^{3}+5^{3}= x^{3}+125$

$(x-5)(x^{2}+5x+25) = (x-5)(x^{2}+5 \cdot x +5^{2}) = x^{3}-5^{3}= x^{3}-125$

Therefore,

$(x+5)(x^{2}-5x+25) - (x-5)(x^{2}+5x+25)$

$= (x^{3} +125) - (x^{3} -125)$

$= x^{3} +125 - x^{3} +125$

= 250

This is not one of the responses.

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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 #21 : Simplifying Algebraic Expressions

Example Question #21 : Simplifying Algebraic Expressions

Example Question #23 : Simplifying Algebraic Expressions

Example Question #24 : Simplifying Algebraic Expressions

Example Question #25 : Simplifying Algebraic Expressions

Example Question #1337 : Problem Solving Questions

Example Question #21 : Simplifying Algebraic Expressions

Example Question #21 : Simplifying Algebraic Expressions

Example Question #1340 : Problem Solving Questions

Example Question #26 : Simplifying Algebraic Expressions

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