Single-Variable Algebra - GED Math

Card 0 of 840

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

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

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

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

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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Question

Simplify

Answer

The first step is to distribute the number outside of the parenthesis to the values inside the parenthesis.

Then in order to simplify you would combine like terms.

.

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