Simplifying, Distributing, and Factoring - GED Math

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Question

Multiply:

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

Answer

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

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Question

Factor:

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

Answer

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

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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Question

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

Answer

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

:

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

:

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

$= 16- 13 \cdot 4 + 36$

:

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

$= 256- 13 \cdot 16 + 36$

:

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

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

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

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Question

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

Answer

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

:

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

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

$= 64 - 66 \cdot 8 + 128$

:

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

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

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

:

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

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

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

:

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

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

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

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

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Question

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

Answer

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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Question

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

Answer

$x^{6} +1$ 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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Question

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

Answer

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.

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Question

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

Answer

$n^{4} + 30n^{2} +225$ 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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Question

A triangle has a base of ft and height of ft. What is the area (in square feet) of the triangle?

Answer

The area of a triangle is: $A= \frac{1}{2}bh$

Use the FOIL Method to simplify.

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

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

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

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

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Question

Express 286 in base five.

Answer

To convert a base ten number to base five, divide the number by five, with the remainder being the digit in the units place; continue, dividing each successive quotient by five and putting the remainder in the next position to the left until the final quotient is less than five.

$286\div 5 = 57 \textrm{ R }\underline{1}$ - 1 is the last digit.

$57 \div 5 = 11 \textrm{ R }\underline{2}$ - 2 is the second-to-last digit.

$11 \div 5 = \underline{2} \textrm{ R }\underline{1}$ - 1 is the third-to-last digit; 2 is the first digit.

286 is equal to $2121_{\textrm{five}}$ .

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Question

Increase by 20%. Which of the following will this be equal to?

Answer

A number increased by 20% is equivalent to 100% of the number plus 20% of the number. This is taking 120% of the number, or, equivalently, multiplying it by 1.2.

Therefore, increased by 20% is 1.2 times this, or

$1.2 \left (8x + 9y \right ) = 1.2 \cdot 8x + 1.2 \cdot 9y = 9.6x + 10.8 y$ .

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Question

Divide:

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

Answer

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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Question

Multiply:

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

Answer

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

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

This product fits the difference of cubes pattern, where :

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

so

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

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

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Question

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

Answer

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

and .

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

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

so . This is the correct choice.

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Question

Simplify the following:

Answer

Group all like terms by their order:

Simplify:

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Question

Simplify the following:

Answer

This can be solved using the FOIL method. The steps are shown below.

Therefore, after reordering, the answer is:

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Question

Decrease by 40%. Which of the following will this be equal to?

Answer

A number decreased by 40% is equivalent to 100% of the number minus 40% of the number. This is taking 60% of the number, or, equivalently, multiplying it by 0.6.

Therefore, decreased by 40% is 0.6 times this, or

$0.6 \left (8t + 72 \right ) =4.8t + 43.2$ .

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Question

Simplify:

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

Answer

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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Question

Factor completely:

$8t^{3} -56t^{2} +3t -21$

Answer

Use the grouping technique, then distribute out the greatest common factor of each group as follows:

$8t^{3} -56t^{2} +3t -21$

$=\left ( 8t^{3} -56t^{2} \right ) +\left (3t -21 \right )$

$=\left ( 8t^{2} \cdot t -8t^{2} \cdot 7 \right ) +\left (3 \cdot t -3 \cdot 7\right )$

$=8t^{2} \left ( t - 7 \right ) + 3\left ( t - 7 \right )$

$=\left (8t^{2} + 3 \right ) \left ( t - 7 \right )$

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Question

Factor completely:

$9t^{2} - 81t$

Answer

The common factor of the terms $9t^{2}$ and can be found as follows:

, and the lesser of the two powers of is ; therefore, $GCF (9t^{2},81t) = 9t$ , their product. Distribute this out:

$9t^{2} - 81t$

$= 9t \cdot t - 9t \cdot 9$

$= 9t \left (t - 9 \right )$

This is as far as we can go with the factoring.

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