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# Divide Polynomials

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

Divide the polynomials.
Use at least two methods as applicable.
Show all work.
Except for ACT and JAMB questions and the like, write your answers using any of the approved formats.
Check your solution using the Division Algorithm

(1.) JAMB Divide $a^{3x} - 26a^{2x} + 156a^x - 216$ by $a^{2x} - 24a^x + 108$

$A.\:\: a^x - 2 \\[3ex] B.\:\: a^x + 2 \\[3ex] C.\:\: a^x - 18 \\[3ex] D.\:\: a^x - 6 \\[3ex]$

$\begin{array}{c|c} & a^x - 2 \\ \hline a^{2x} - 24a^x + 108 & a^{3x} - 26a^{2x} + 156a^x - 216 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &a^{3x} - 24a^{2x} + 108a^x~~~~~~~~~~ \\ \hline &~~~~~~~-2a^{2x} + 48a^x - 216 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~-2a^{2x} + 48a^x - 216 \\ \hline &~~~~~~~0 \\ \end{array} \\[5ex] Quotient = a^x - 2$
(2.)

$f(3)$
This means that we have to substitute $3$ for $x$

$f(x) = -3x + 7 \\[3ex] f(3) = -3(3) + 7 \\[3ex] f(3) = -9 + 7 \\[3ex] f(3) = -2 \\[3ex]$ $f(-5)$
This means that we have to substitute $-5$ for $x$

$f(x) = -3x + 7 \\[3ex] f(-5) = -3(-5) + 7 \\[3ex] f(-5) = 15 + 7 \\[3ex] f(-5) = 22$
(3.)

$g(3)$
This means that we have to substitute $3$ for $x$

$g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 3 - \dfrac{3}{4} \\[5ex] = \dfrac{2}{1} - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{8 - 3}{4} \\[5ex] = \dfrac{5}{4} \\[5ex]$ $g(4)$
This means that we have to substitute $4$ for $x$

$g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 4 - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{32}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{32 - 9}{12} \\[5ex] = \dfrac{23}{12} \\[5ex]$ $g\left(-\dfrac{1}{2}\right)$
This means that we have to substitute $-\dfrac{1}{2}$ for $x$

$g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(4) = \dfrac{2}{3} * -\dfrac{1}{2} - \dfrac{3}{4} \\[5ex] = -\dfrac{1}{3} - \dfrac{3}{4} \\[5ex] = -\dfrac{4}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{-4 - 9}{12} \\[5ex] = -\dfrac{13}{12}$
(4.)

$f(-a)$
This means that we have to substitute $(-a)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(-a) = (-a)^2 - 3(-a) \\[3ex] = a^2 + 3a \\[3ex] = a(a + 3) \\[3ex]$ $f(a - 3)$
This means that we have to substitute $(a - 3)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = a^2 - 3a - 3a + 9 - 3a + 9 \\[3ex] = a^2 - 9a + 18 \\[3ex] = (a - 3)(a - 6) \\[3ex]$ OR

$f(x) = x^2 - 3x \\[3ex] f(a - 3) = (a - 3)^2 - 3(a - 3) \\[3ex] = (a - 3)(a - 3) - 3a + 9 \\[3ex] = (a - 3)(a - 3) - 3(a - 3) \\[3ex] = (a - 3)[(a - 3) - 3] \\[3ex] = (a - 3)[a - 3 - 3] \\[3ex] = (a - 3)(a -6) \\[3ex]$ $f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$f(x) = x^2 - 3x \\[3ex] f(a + h) = (a + h)^2 - 3(a + h) \\[3ex] = (a + h)(a + h) - 3(a + h) \\[3ex] = (a + h)[(a + h) - 3] \\[3ex] = (a + h)(a + h - 3)$
(5.)

$p(3)$
This means that we have to substitute $3$ for $m$

$p(m) = m^3 \\[3ex] p(3) = 3^3 \\[3ex] = 27 \\[3ex]$ $p(-2)$
This means that we have to substitute $(-2)$ for $m$

$p(m) = m^3 \\[3ex] p(-2) = (-2)^3 \\[3ex] = -8 \\[3ex]$ $p(-m)$
This means that we have to substitute $(-m)$ for $m$

$p(m) = m^3 \\[3ex] p(3) = (-m)^3 \\[3ex] = (-m)(-m)(-m) \\[3ex] = -m^3 \\[3ex]$ $p(3y)$
This means that we have to substitute $(3y)$ for $m$

$p(m) = m^3 \\[3ex] p(3y) = (3y)^3 \\[3ex] = 27y^3 \\[3ex]$ $p(2 + h)$
This means that we have to substitute $(2 + h)$ for $m$

$p(m) = m^3 \\[3ex] p(2 + h) = (2 + h)^3 \\[3ex] = (2 + h)(2 + h)(2 + h) \\[3ex] = (4 + 2h + 2h + h^2)(2 + h) \\[3ex] = (h^2 + 4h + 4)(2 + h) \\[3ex] = (2h^2 + h^3 + 8h + 4h^2 + 8 + 4h) \\[3ex] = h^3 + 6h^2 + 12h + 8$
(6.)

$p(0)$
This means that we have to substitute $0$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(0) = -1 * 0^2 - 5(0) + 8 \\[3ex] p(0) = -1 * 0 - 0 + 8 \\[3ex] p(0) = 0 - 0 + 8 \\[3ex] p(0) = 8 \\[3ex]$ $p(7)$
This means that we have to substitute $7$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(7) = -1 * 7^2 - 5(7) + 8 \\[3ex] p(7) = -1 * 49 - 35 + 8 \\[3ex] p(7) = -49 - 35 + 8 \\[3ex] p(7) = -76 \\[3ex]$ $p(-7)$
This means that we have to substitute $-7$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(-7) = -1 * (-7)^2 - 5(-7) + 8 \\[3ex] p(-7) = -1 * 49 + 35 + 8 \\[3ex] p(-7) = -49 + 35 + 8 \\[3ex] p(-7) = -6 \\[3ex]$ $p(g)$
This means that we have to substitute $g$ for $x$

$p(x) = -x^2 - 5x + 8 \\[3ex] p(g) = -g^2 - 5g + 8$
(7.)

$f(a)$
This means that we have to substitute $a$ for $x$

$f(x) = -4x + 7 \\[3ex] f(a) = -4a + 7 \\[3ex]$
$f(a + 2)$
This means that we have to substitute $(a + 2)$ for $x$

$f(x) = -4x + 7 \\[3ex] f(a + 2) = -4(a + 2) + 7 \\[3ex] = -4a - 8 + 7 \\[3ex] = -4a - 1 \\[3ex]$
$f(a + h)$
This means that we have to substitute $(a + h)$ for $x$

$f(x) = -4x + 7 \\[3ex] f(a + h) = -4(a + h) + 7 \\[3ex] = -4a - 4h + 7$
(8.)

$f(x) = -4x^3 - 4x^2 \\[3ex] f(-4) \rightarrow x = -4 \\[3ex] f(-4) = -4[(-4)^3] - 4(-4)^2 \\[3ex] f(-4) = -4(-64) - 4(16) \\[3ex] f(-4) = 256 - 64 \\[3ex] f(-4) = 192$