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GED Math : Simplifying, Distributing, and Factoring

Study concepts, example questions & explanations for GED Math

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Example Questions

Example Question #1 : Algebra

Multiply:

$\small -x(4x+2)=$

Possible Answers:

$\small 4x^2+2x$

$\small -4x^2-2x$

$\small -4x^2+2x$

$\small 4x^2-2x$

Correct answer:

$\small -4x^2-2x$

Explanation:

$\small -x(4x+2)=(-x)(4x)+(-x)(2)=-4x^2+(-2x)=-4x^2-2x$

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Example Question #1 : Simplifying, Distributing, And Factoring

Factor:

$\small x^2+5x+4$

Possible Answers:

$\small (x+5)(x-1)$

$\small (x-4)(x-1)$

$\small (x+4)(x+1)$

$\small (x-5)(x+1)$

Correct answer:

$\small (x+4)(x+1)$

Explanation:

$\small x^2+5x+4=(x+a)(x+b)$

where

$\small a+b=5\ and\ ab=4$

The numbers $\small 4$ and $\small 1$ fit those criteria. Therefore,

$\small x^2+5x+4=(x+1)(x+4)$

You can double check the answer using the FOIL method

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Example Question #3 : Algebra

Which of the following is not a prime factor of $y^{4}- 81$ ?

Possible Answers:

y + 3

y - 3

$y^{2}+9$

$y^{2}-9$

Correct answer:

$y^{2}-9$

Explanation:

Factor $y^{4}- 81$ all the way to its prime factorization.

$y^{4}- 81$ can be factored as the difference of two perfect square terms as follows:

$y^{4}- 81$

$=\left ( y^{2} \right )^{2} - 9^{2}$

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

$y^{2} + 9$ is a factor, and, as the sum of squares, it is a prime. $y^{2} - 9$ is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

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

$=\left ( y^{2} + 9 \right )\left ( y^{2} - 3^{2} \right )$

$=\left ( y^{2} + 9 \right )\left (y+3 \right )\left ( y-3\right )$

Therefore, all of the given polynomials are factors of $y^{4}- 81$ , but $y^{2} - 9$ is the correct choice, as it is not a prime factor.

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Example Question #2 : Simplifying, Distributing, And Factoring

Which of the following is a prime factor of $n^{4} + 30n^{2} +225$ ?

Possible Answers:

$n^{2}+9$

$n^{2}+25$

$n^{2}+75$

$n^{2}+15$

Correct answer:

$n^{2}+15$

Explanation:

$n^{4} + 30n^{2} +225$ can be seen to fit the pattern

$A ^{2} + 2AB + B^{2}$ :

$n^{4} + 30n^{2} +225 = \left (n^{2} \right ) ^{2}+ 2 \cdot n^{2} \cdot 15 + 15 ^{2}$

where $A = n^{2} , B = 15$

$A ^{2}+2AB + B^{2}$ can be factored as $(A+B ) ^{2}$ , so

$n^{4} + 30n^{2} +225 = \left (n^{2} \right ) ^{2}+ 2 \cdot n^{2} \cdot 15 + 15 ^{2} = (n^{2}+15)^{2}$ .

$n^{2}+15$ does not fit into any factorization pattern, so it is prime, and the above is the complete factorization of the polynomial. Therefore, $n^{2}+15$ is the correct choice.

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Example Question #5 : Algebra

Divide:

$\left ( 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x \right ) \div 6x$

Possible Answers:

$6x ^{4} + 3x ^{2} +x + 30$

$6x ^{4} + 3x ^{2} + 30$

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

$2x ^{4} +3x ^{2} + 6$

Correct answer:

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

Explanation:

Divide termwise:

$\left ( 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x \right ) \div 6x$

$= \frac{ 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x }{6x}$

$= \frac{ 12 x ^{5} }{6x} + \frac{ 9x^{3} }{6x} + \frac{ 6x^{2} }{6x}+ \frac{ 36 x }{6x}$

$= \frac{ 12 }{6} x ^{5-1} + \frac{ 9 }{6 }x ^{3-1} + \frac{ 6 }{6}x^{2-1}+ \frac{ 36 }{6 }$

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

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Example Question #5 : Algebra

Multiply:

$\left (100Y^{2}+ 110Y + 121 \right )(10Y-11)$

Possible Answers:

$1,000Y^{3}- 1,100Y^{2}+1,210Y - 1,331$

$1,000Y^{3}+ 1,100Y^{2}-1,210Y - 1,331$

$1,000Y^{3}- 3,300Y^{2}+3,630Y - 1,331$

$1,000Y^{3}- 1,331$

Correct answer:

$1,000Y^{3}- 1,331$

Explanation:

$\left (100Y^{2}+ 110Y + 121 \right )(10Y-11)$

$=\left [\left (10Y \right) ^{2}+ 10Y \cdot 11 + 11^{2} \right] (10Y-11)$

This product fits the difference of cubes pattern, where A = 10Y, B = 11 :

$(A^{2}+AB +B^{2})(A-B) = A^{3}-B^{3}$

$\left [\left (10Y \right) ^{2}+ 10Y \cdot 11 + 11^{2} \right] (10Y-11)$

$= \left ( 10Y\right )^{3} -11^{3} = 1,000Y^{3}-1,331$

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Example Question #6 : Algebra

Give the value of that makes the polynomial $121x^{2}- Nx + 49$ the square of a linear binomial.

Possible Answers:

N = 231

N = 198

N = 154

N = 77

Correct answer:

N = 154

Explanation:

A quadratic trinomial is a perfect square if and only if takes the form

$A^{2} \pm 2AB + B^{2}$ for some values of and .

$121x^{2}- Nx + 49= (11x)^{2} - Nx + 7^{2}$ , so

A = 11x and B = 7 .

For $121x^{2}- Nx + 49$ to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 11x \cdot 7 = 154x$ ,

so N = 154 . This is the correct choice.

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Example Question #1 : Algebra

Which of the following is a factor of the polynomial $x^{4}-13x^{2}+ 36$ ?

Possible Answers:

x - 4

x-2

x-9

x-1

Correct answer:

x-2

Explanation:

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that x-a is a factor of polynomial P(x) if and only if P(a)= 0 . We substitute 1, 2, 4, and 9 for in the polynomial to identify the factor.

x=1 :

$1^{4}-13 \cdot 1^{2}+ 36$

= 1 - 13 + 36

= 24

x=2 :

$2^{4}-13 \cdot 2^{2}+ 36$

$= 16- 13 \cdot 4 + 36$

= 16 -52 + 36

= 0

x =4 :

$4^{4}-13 \cdot 4^{2}+ 36$

$= 256- 13 \cdot 16 + 36$

= 256 -208 +36

= 84

x=9 :

$9^{4}-13 \cdot 9^{2}+ 36$

$= 6,561- 13 \cdot 81 + 36$

= 6,561- 1,053+ 36

= 5,544

Only x = 2 makes the polynomial equal to 0, so among the choices, only x-2 is a factor.

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Example Question #3 : Simplifying, Distributing, And Factoring

Which of the following is a prime factor of $x^{6} +1$ ?

Possible Answers:

$x^{2}-x+1$

$x^{2}+1$

x+1

$x^{2}+x+1$

Correct answer:

$x^{2}+1$

Explanation:

$x^{6} +1$ is the sum of two cubes:

$x^{6} +1 =\left ( x^{2} \right ) ^{3}+1^{3}$

As such, it can be factored using the pattern

$A^{3}+ B^{3}= (A+B)(A^{2}-AB+B^{2})$

where $A = x^{2}, B = 1$ ;

$x^{6} +1 =\left ( x^{2} \right ) ^{3}+1^{3}$

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

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

The first factor,as the sum of squares, is a prime.

We try to factor the second by noting that it is "quadratic-style" based on $x^{2}$ . and can be written as

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

we seek to factor it as $(x^{2}+\; \; \; )(x^{2}+\; \; \; )$

We want a pair of integers whose product is 1 and whose sum is . These integers do not exist, so $x^{4}-x^{2} + 1$ is a prime.

$(x^{2}+1)(x^{4}-x^{2} + 1)$ is the prime factorization and the correct response is $x^{2}+1$ .

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Example Question #563 : Ged Math

Which of the following is a factor of the polynomial $x^{6} - 66x^{3} + 128$

Possible Answers:

x+4

x+2

x-4

x-2

Correct answer:

x-4

Explanation:

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that x-a is a factor of polynomial P(x) if and only if P(a)= 0 . We substitute $\pm2$ and $\pm 4$ for in the polynomial to identify the factor.

x = 2 :

$x^{6} - 66x^{3} + 128$

$2^{6} - 66 \cdot 2^{3} + 128$

$= 64 - 66 \cdot 8 + 128$

= 64 - 528 + 128

= -336

x = -2 :

$x^{6} - 66x^{3} + 128$

$\left (-2 \right )^{6} - 66 \cdot \left (-2 \right )^{3} + 128$

$= 64 - 66 \cdot\left ( -8 \right )+ 128$

= 64 + 528 + 128

= 720

x = 4 :

$x^{6} - 66x^{3} + 128$

$4^{6} - 66 \cdot 4^{3} + 128$

$= 4,096- 66 \cdot 64 + 128$

= 4,096- 4,224 + 128

= 0

x = -4 :

$x^{6} - 66x^{3} + 128$

$\left (-4 \right )^{6} - 66 \cdot \left (-4 \right )^{3} + 128$

$= 4,096- 66 \cdot\left ( -64 \right )+ 128$

= 4,096+4,224 + 128

= 8,448

Only x = 4 makes the polynomial equal to 0, so of the four choices, only x - 4 is a factor of the polynomial.

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GED Math : Simplifying, Distributing, and Factoring

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #1 : Algebra

Example Question #1 : Simplifying, Distributing, And Factoring

Example Question #3 : Algebra

Example Question #2 : Simplifying, Distributing, And Factoring

Example Question #5 : Algebra

Example Question #5 : Algebra

Example Question #6 : Algebra

Example Question #1 : Algebra

Example Question #3 : Simplifying, Distributing, And Factoring

Example Question #563 : Ged Math

All GED Math Resources

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