SAT Math - Operations with Polynomials

Divide:

$\left ( 40x^{3} - 24x^{2}+ 20x- 64\right ) \div 8 x^{2}$

$5x-3+ \frac{ 5}{ 2x } - \frac{8}{x^{2} }$

$5x-3+ \frac{ 5}{ 2x^{2} } - \frac{8}{x^{3} }$

$5x^{2} -3+ \frac{ 5}{ 2x^{2} } - \frac{8}{x^{3} }$

$5x-3 - \frac{8}{x}+ \frac{ 5}{ 2x^{2} }$

$5x-3 - \frac{8}{x^{2}}+ \frac{ 5}{ 2x^{3} }$

Explanation

Divide termwise:

$\left ( 40x^{3} - 24x^{2}+ 20x- 64\right ) \div 8 x^{2}$

$= \frac{40x^{3} - 24x^{2}+ 20x- 64}{ 8 x^{2}}$

$= \frac{40x^{3} }{ 8 x^{2}}- \frac{ 24x^{2}}{ 8 x^{2}}+ \frac{ 20x}{ 8 x^{2}}- \frac{ 64}{ 8 x^{2}}$

$= \frac{40 }{ 8}x^{3-2}- \frac{ 24}{ 8 }+ \frac{ 20x}{ 8 } \cdot \frac{1}{x^{2-1}}- \frac{ 64}{8 }\cdot \frac{1}{x^{2} }$

$=5x-3+ \frac{ 5}{ 2 } \cdot \frac{1}{x}- 8 \cdot \frac{1}{x^{2} }$

$=5x-3+ \frac{ 5}{ 2x } - \frac{8}{x^{2} }$

Which of the following is a prime factor of $n^{4} - 32n^{2} +256$ ?

None of the other responses gives a correct answer.

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

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

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

$n^{2} + 4n + 16$

Explanation

$n^{4} - 32n^{2} +256$ can be seen to fit the pattern

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

$n^{4} - 32n^{2} +256 = \left (n^{2} \right ) ^{2}- 2 \cdot n^{2} \cdot 16 + 16 ^{2}$

where $A = n^{2} , B = 16$

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

$n^{4} - 32n^{2} +256 = \left (n^{2} \right ) ^{2}- 2 \cdot n^{2} \cdot 16 + 16 ^{2} =\left ( n ^{2} - 16 \right )^{2}$

$n^{2} - 16 = n^{2} - 4^{2}$ , making this the difference of squares, so it can be factored as follows:

$n^{2} - 16 = n^{2} - 4^{2} = (n + 4) (n - 4)$

Therefore,

$n^{4} - 32n^{2} +256 = \left ( n ^{2} - 16 \right )^{2} = (n+4) ^{2} (n-4) ^{2}$

The polynomial has only two prime factors, each squared, neither of which appear among the choices.

Factor:

$27N^{3}-1,331$

$(3N - 11)(9N^{2}+33N+121)$

$(3N - 11)(9N^{2}+121)$

$(3N - 11)^{3}$

$(3N - 11)(3N+11)^{2}$

The polynomial is prime.

Explanation

$27N^{3}-1,331$ can be rewritten as $(3N)^{3} - 11^{3}$ and is therefore the difference of two cubes. As such, it can be factored using the pattern

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

where .

$27N^{3}-1,331$

$=(3N)^{3} - 11^{3}$

$= (3N-11)[(3N)^{2}+3N\cdot 11 +11^{2}]$

$= (3N-11)(9N^{2}+33N +121)$

Factor completely:

$t^{3} -225$

The polynomial is prime.

$(t + 15)(t^{2}-15t + 15)$

$(t - 15)(t^{2}-15t - 15)$

$(t + 1)(t^{2}-15t + 225)$

$(t -1)(t^{2}-15t - 225)$

Explanation

Since the first term is a perfect cube, the factoring pattern we are looking to take advantage of is the difference of cubes pattern. However, 225 is not a perfect cube of an integer $\left (6^{3} = 216 < 225 < 7^{3} = 343 \right )$ , so the factoring pattern cannot be applied. No other pattern fits, so the polynomial is a prime.

Factor completely:

$t^{3} + 512$

$(t+ 8) (t^{2}-8t+64)$

The polynomial is prime.

$(t- 8) (t^{2}-8t-64)$

$(t+ 8) (t-8)^{2}$

$(t+ 8) ^{3}$

Explanation

Since both terms are perfect cubes $\left (512= 8^{3} \right )$ , the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is

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

We substitute for and 8 for :

$t^{3} + 512= t^{3} + 8^{3}$

$= (t+ 8) (t^{2}-t \cdot 8+8^{2})$

$= (t+ 8) (t^{2}-8t+64)$

Subtract the expressions below.

$\left (\frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2} \right )-\left (\frac{1}{9}x^{2}-\frac{4}{3}xy+y^{2} \right )$

$\frac{2}{9}x^{2}+\frac{3}{2}xy-\frac{9}{4}y^{2}$

$\frac{2}{9}x^{2}+\frac{7}{6}xy-\frac{1}{4}y^{2}$

$\frac{2}{9}x^{2}+\frac{3}{2}xy-\frac{5}{4}y^{2}$

$\frac{1}{27}x^{2}-\frac{2}{9}xy-\frac{5}{4}y^{2}$

None of the other answers are correct.

Explanation

$\left (\frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2} \right )-\left (\frac{1}{9}x^{2}-\frac{4}{3}xy+y^{2} \right )$

Since we are only adding and subtracting (there is no multiplication or division), we can remove the parentheses.

$\frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2}-\left \frac{1}{9}x^{2}+\frac{4}{3}xy-y^{2} \right$

Regroup the expression so that like variables are together. Remember to carry positive and negative signs.

$(\frac{1}{3}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{4}{3}xy)-(\frac{5}{4}y^{2}-y^{2} \right)$

For all fractional terms, find the least common multiple in order to add and subtract the fractions.

$(\frac{3}{9}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{8}{6}xy)-(\frac{5}{4}y^{2}-\frac{4}{4}y^{2} \right)$

Combine like terms and simplify.

$\frac{2}{9}x^{2}+\frac{9}{6}xy-\frac{9}{4}y^{2}$

$\frac{2}{9}x^{2}+\frac{3}{2}xy-\frac{9}{4}y^{2}$

Divide by .

$3x+5+\frac{2x+13}{x^2-1}$

$2x+13+\frac{3x+5}{x^2-1}$

$x^2-1-\frac{3x+5}{2x+13}$

$3x+5+\frac{x^2+1}{2x-13}$

$3x+5-\frac{2x+13}{x^2-1}$

Explanation

First, set up the division as the following:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && & & & \ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}$

Look at the leading term in the divisor and in the dividend. Divide by gives ; therefore, put on the top:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}$

Then take that and multiply it by the divisor, , to get . Place that under the division sign:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \ & && 3x^3 & -3x \end{array}$

Subtract the dividend by that same and place the result at the bottom. The new result is , which is the new dividend.

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \ & && 3x^3 & -3x \ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}$

Now, is the new leading term of the dividend. Dividing by gives 5. Therefore, put 5 on top:

$\begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \ & && 3x^3 & -3x \ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}$