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GED Math : Single-Variable Algebra

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Example Questions

Example Question #21 : Algebra

Simplify completely:

$\frac{3xy \cdot 7xy}{12y^{3}}$

Possible Answers:

$\frac{7 x }{4y^{2} }$

$\frac{7 x^{2}y }{4 }$

$\frac{7 x^{2} }{4y^{2} }$

$\frac{7 x^{2} }{4y }$

Correct answer:

$\frac{7 x^{2} }{4y }$

Explanation:

$\frac{3xy \cdot 7xy}{12y^{3}}$

$= \frac{3 \cdot 7 \cdot x \cdot x \cdot y \cdot y}{12y^{3}}$

$= \frac{21 x^{2} y^{2} }{12y^{3}}$

$= \frac{21 x^{2} }{12y^{3-2}}$

$= \frac{21 x^{2} }{12y }$

$= \frac{21 x^{2} \div 3 }{12y \div 3}$

$= \frac{7 x^{2} }{4y }$

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Example Question #21 : Algebra

Subtract $x^{2} - 5$ from $8x ^{2} -7x + 4$ .

Possible Answers:

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

$8x ^{2} -8x -1$

$8x ^{2} -8x + 9$

$7x ^{2} -7x -1$

Correct answer:

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

Explanation:

$(8x ^{2} -7x + 4) -(x^{2} - 5)$

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

$= 8x ^{2} -1x^{2} -7x + 4 + 5$

$=\left ( 8-1 \right )x ^{2} -7x +\left ( 4 + 5 \right )$

$=7x ^{2} -7x +9$

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Example Question #22 : Algebra

Add $x^{2} - 5$ to $8x ^{2} -7x + 4$ .

Possible Answers:

$8x ^{2} -6x -1$

$8x ^{2} -6x -9$

$9x ^{2} -7x -1$

$9x ^{2} -7x -9$

Correct answer:

$9x ^{2} -7x -1$

Explanation:

$(8x ^{2} -7x + 4) +(x^{2} - 5)$

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

$=\left ( 8+1 \right ) x ^{2} -7x +\left ( 4 - 5 \right )$

$=9x ^{2} -7x -1$

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Example Question #23 : Algebra

Factor completely:

$5y^{3}- 20y^{2}- 105y$

Possible Answers:

5y (y-7)(y + 3)

5y (y-21)(y + 1)

5y (y-3)(y + 7)

5y (y-1)(y +2 1)

Correct answer:

5y (y-7)(y + 3)

Explanation:

First, factor out the greatest common factor of the terms, which is :

$5y^{3}- 20y^{2}- 105y$

$= 5y \cdot y^{2}- 5y \cdot 4y- 5y \cdot 21$

$= 5y \left (y^{2}- 4y- 21 \right )$

The quadratic trinomial can be factored as (x+A)(x+ B) where A = -21 and A + B = -4 ; by trial and error we find that the numbers chosen are A = -7, B = 3 , so

$5y \left (y^{2}- 4y- 21 \right )$

= 5y (y-7)(y + 3)

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Example Question #21 : Single Variable Algebra

Simplify:

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

Possible Answers:

$\frac{9}{t^{3}}$

$\frac{27}{t^{3}}$

$\frac{t^{3}}{27}$

$9}{t^{3}$

Correct answer:

$\frac{27}{t^{3}}$

Explanation:

Apply the power of a quotient rule:

$\left (\frac{3}{t} \right )^{3} = \frac{3^{3}}{t^{3}} = \frac{27}{t^{3}}$

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Example Question #22 : Algebra

Factor completely:

$x^{3} + 121x$

Possible Answers:

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

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

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

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

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 #22 : Single Variable Algebra

Factor completely:

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

Possible Answers:

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

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

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

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

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 #23 : Algebra

Factor completely:

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

Possible Answers:

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

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

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

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

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

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 #24 : 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 #25 : Algebra

Factor completely:

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

Possible Answers:

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

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

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

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

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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GED Math : Single-Variable Algebra

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #21 : Algebra

Example Question #21 : Algebra

Example Question #22 : Algebra

Example Question #23 : Algebra

Example Question #21 : Single Variable Algebra

Example Question #22 : Algebra

Example Question #22 : Single Variable Algebra

Example Question #23 : Algebra

Example Question #24 : Algebra

Example Question #25 : Algebra

All GED Math Resources

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