GED Math - Simplifying, Distributing, and Factoring

Simplify:

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

$\frac{x^{8}}{4}$

$\frac{1}{4x^{8}}$

$\frac{x^{8}}{256}$

$\frac{1}{256x^{8}}$

Explanation

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

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

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

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

$= \frac{ x^{4 \cdot 2} }{2^{2}}$

$= \frac{ x^{8} }{4}$

Simplify completely:

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

Simplify:

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

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

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

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

$9}{t^{3}$

Explanation

Apply the power of a quotient rule:

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

Simplify:

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

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

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

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

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

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:

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

Multiply:

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

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

$\small 4x^2-2x$

$\small 4x^2+2x$

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

Explanation

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

Which of the following is a prime factor of $n^{4} - 16n^{2} - 225$ ?

$n^{2}+9$

$n^{2}+25$

$n^{2}+3$

$n^{2}+75$

Explanation

This can be most easily solved by first substituting for $n^{2}$ , and, subsequently, $t^{2}$ for $n^{4}$ :

$n^{4} - 16n^{2} - 225$

$= t^{2} - 16t - 225$

This becomes quadratic in the new variable, and can be factored as

$\left (t +; ; \right )\left (t +; ; \right )$ ,

filling out the blanks with two numbers whose sum is and whose product is . Through some trial and error, the numbers can be seen to be .

Therefore, after factoring and substituting back,

$n^{4} - 16n^{2} - 225$

$= t^{2} - 16t - 225$

$= (n^{2}+9)(n^{2}-25)$

The first factor, the sum of squares, is prime. The second factors as the difference of squares, so the final factorization is

$(n^{2}+9)(n+5)(n-5)$ .

Of the choices given, $n^{2}+9$ is correct.

Divide:

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

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

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

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

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

Explanation

Divide termwise:

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

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

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

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

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

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

$n^{2}+15$

$n^{2}+25$

$n^{2}+9$

$n^{2}+75$

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.