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GED Math : GED Math

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4 Diagnostic Tests 263 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #26 : Algebra

Factor completely:

$x^{3} + 121x$

Possible Answers:

$\left (x ^{2} + 11 \right ) (x+ 11)$

$x (x^{2}+ 121)$

$x^{2} (x+ 121)$

$x \left ( x + 11\right )^{2}$

Correct answer:

$x (x^{2}+ 121)$

Explanation:

is a common factor of both terms, so factor it out:

$x^{3} + 121x$

$= x \cdot x ^{2} +x \cdot 121$

$= x \left ( x ^{2} + 121 \right )$

$= x ^{2} + 121$ cannot be factored, so this is the complete factorization.

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Example Question #27 : Algebra

Factor completely:

$12x^{2} + 19x + 4$

Possible Answers:

$2 \left (3x + 2 \right )\left (2x+ 1 \right )$

$2\left (x + 2 \right )\left (6x+ 1 \right )$

$\left (3x + 4 \right )\left (4x+ 1 \right )$

$4 \left (3x + 1 \right )\left (x+1 \right )$

Correct answer:

$\left (3x + 4 \right )\left (4x+ 1 \right )$

Explanation:

First, we find two integers whose sum is 19 and whose product is $12 \times 4 = 48$ . Through trial and error we find these integers are 3 and 16. We use these numbers to split the middle term, then we factor using the grouping method:

$12x^{2} + 19x + 4$

$= 12x^{2}+ 3x + 16x + 4$

$= \left (12x^{2}+ 3x \right )+ \left (16x + 4 \right )$

$= \left (3x \cdot 4x+ 3x \cdot 1 \right )+ \left (4 \cdot 4x + 4 \cdot 1 \right )$

$= 3x \left (4x+ 1 \right )+ 4\left (4x+ 1 \right )$

$= \left (3x + 4 \right )\left (4x+ 1 \right )$

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Example Question #28 : Algebra

Factor completely:

$5x ^{3}+ 25x^{2} - 7x - 35$

Possible Answers:

$5 \left (x ^{2} +1 \right ) \left ( x - 7 \right )$

$\left (5x ^{2} - 7 \right ) \left ( x + 5 \right )$

$\left (5x ^{2} + 7 \right ) \left ( x - 5 \right )$

$\left (5x ^{2} - 1 \right ) \left ( x + 35 \right )$

$\left (5x ^{2} +1 \right ) \left ( x - 35 \right )$

Correct answer:

$\left (5x ^{2} - 7 \right ) \left ( x + 5 \right )$

Explanation:

Factor by grouping as follows:

$5x ^{3}+ 25x^{2} - 7x - 35$

$=\left (5x ^{3}+ 25x^{2} \right )- \left (7x + 35 \right )$

$=\left (5x ^{2} \cdot x + 5x^{2} \cdot 5 \right )- \left (7 \cdot x + 7 \cdot 5 \right )$

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

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

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Example Question #29 : Algebra

Factor completely:

$x ^{3} + 7x ^{2} - 9x - 63$

Possible Answers:

$\left (x^{2}-7 \right )\left ( x+ 9 \right )$

$\left (x^{2}+ 9\right )\left ( x-7\right )$

$\left ( x + 3 \right )\left ( x - 3 \right ) \left ( x + 7 \right )$

$\left (x^{2}+ 7 \right )\left ( x-9\right )$

Correct answer:

$\left ( x + 3 \right )\left ( x - 3 \right ) \left ( x + 7 \right )$

Explanation:

Factor by grouping as follows:

$x ^{3} + 7x ^{2} - 9x - 63$

$= \left (x ^{3} + 7x ^{2} \right )- \left (9x + 63 \right )$

$= \left (x ^{2} \cdot x + x ^{2} \cdot 7 \right )- \left (9 \cdot x + 9 \cdot 7\right )$

$= x ^{2} \left ( x + 7 \right )- 9 \left ( x + 7\right )$

$=\left ( x ^{2}- 9 \right ) \left ( x + 7 \right )$

The first factor is the difference of squares, so further factoring can be done:

$=\left ( x ^{2}- 3 ^{2} \right ) \left ( x + 7 \right )$

$=\left ( x + 3 \right )\left ( x - 3 \right ) \left ( x + 7 \right )$

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Example Question #30 : Algebra

Factor completely:

$49x^{2} - 70x+ 25$

Possible Answers:

(7x + 1)(7x - 25)

(7x - 1)(7x - 25)

$(7x - 5)^{2}$

(7x+5)(7x - 5)

Correct answer:

$(7x - 5)^{2}$

Explanation:

The polynomial fits the perfect square pattern:

$49x^{2} - 70x+ 25$

$\left (7x \right )^{2} - 2 \cdot 7x \cdot 5 + 5^{2}$

This can be factored using the pattern

$A^{2} - 2AB + B ^{2} = (A+B)^{2}$

with A = 7x, B = 5 :

$\left (7x \right )^{2} - 2 \cdot 7x \cdot 5 + 5^{2} = (7x - 5)^{2}$

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Example Question #31 : Single Variable Algebra

Factor completely:

$36x^{2} - 49y^{2}$

Possible Answers:

$\left (6x - 7y \right ) ^{2}$

$\left (12x + 7y \right )\left (3x - 7y \right )$

$\left (6x + 7y \right )\left (6x - 7y \right )$

$\left (3x + 7y \right )\left (12x - 7y \right )$

Correct answer:

$\left (6x + 7y \right )\left (6x - 7y \right )$

Explanation:

The polynomial is the difference of squares and can be factored using the pattern

$A^{2} - B ^{2} = (A+B)(A-B)$

where

A = 6x, B = 7y

as seen here:

$36x^{2} - 49y^{2}$

$= 6^{2}x^{2} - 7^{2}y^{2}$

$= \left (6 x \right )^{2} - \left (7 y \right )^{2}$

$= \left (6x + 7y \right )\left (6x - 7y \right )$

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Example Question #32 : Single Variable Algebra

Which of the following is a factor of the polynomial $8x ^{3}y - 20x ^{2}y ^{2}$ ?

Possible Answers:

2- 5xy

$2x ^{2}- 5y ^{2}$

2x - 5y

2xy - 5

Correct answer:

2x - 5y

Explanation:

The greatest common factor of the two terms is the monomial term $4x ^{2}y$ , so factor it out:

$8x ^{3}y - 20x ^{2}y ^{2}$

$= 4x ^{2}y \cdot 2x - 4x ^{2}y \cdot 5y$

$= 4x ^{2}y \left (2x - 5y \right )$

Of the four choices, 2x - 5y is correct.

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Example Question #33 : Single Variable Algebra

Which of the following is a factor of the polynomial $8x ^{3}y - 20x ^{2}y ^{2}$ ?

Possible Answers:

2x - 5y

$2x ^{2}- 5y ^{2}$

2xy - 5

2- 5xy

Correct answer:

2x - 5y

Explanation:

The greatest common factor of the two terms is the monomial term $4x ^{2}y$ , so factor it out:

$8x ^{3}y - 20x ^{2}y ^{2}$

$= 4x ^{2}y \cdot 2x - 4x ^{2}y \cdot 5y$

$= 4x ^{2}y \left (2x - 5y \right )$

Of the four choices, 2x - 5y is correct.

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Example Question #581 : Ged Math

Simplify:

$\left (\frac{5}{t} \right )^{-3}$

Possible Answers:

$-\frac{125}{t^{3}}$

$-\frac{1 5}{t^{3}}$

$\frac{t^{3}}{125}$

$\frac{t^{3}}{1 5}$

Correct answer:

$\frac{t^{3}}{125}$

Explanation:

Raise a fraction to a negative power by raising its reciprocal to the power of the absolute value of the exponent. Then apply the power of a quotient rule:

$\left (\frac{5}{t} \right )^{-3} = \left (\frac{t}{5} \right )^{ 3} = \frac{t^{ 3}}{5^{ 3}} = \frac{t^{ 3}}{125}$

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Example Question #582 : Ged Math

Simplify:

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

Possible Answers:

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

$\frac{1}{64t^{6}}$

$\frac{4}{ t^{6}}$

$-\frac{12}{ t^{6}}$

Correct answer:

$\frac{1}{64t^{6}}$

Explanation:

To raise a number to a negative exponent, raise it to the absolute value of that exponent, then take its reciprocal. We do this, then apply the various properties of exponents:

$\left (4 t^{2} \right )^{-3} = \frac{1}{\left (4 t^{2} \right )^{3} } = \frac{1}{4^{3} \left ( t^{2} \right )^{3} } =\frac{1}{64 t^{2 \cdot 3} } =\frac{1}{64 t^{6} }$

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GED Math : GED Math

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #26 : Algebra

Example Question #27 : Algebra

Example Question #28 : Algebra

Example Question #29 : Algebra

Example Question #30 : Algebra

Example Question #31 : Single Variable Algebra

Example Question #32 : Single Variable Algebra

Example Question #33 : Single Variable Algebra

Example Question #581 : Ged Math

Example Question #582 : Ged Math

All GED Math Resources

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