Solve for \(x\): \({x}^{3}+{x}^{2}-5x+3=0\)
\begin{align*}
\text{Let } a(x) &= {x}^{3}+{x}^{2}-5x+3 \\
a(1) &= (1)^{3}+(1)^{2}-5(1)+3 \\
&= 0 \\
\therefore a(x) &= (x-1)(x^{2} + 2x - 3) \\
&= (x-1)(x+3)(x-1) \\
&= (x-1)^{2}(x+3) \\
\therefore 0 &= (x-1)^{2}(x+3) \\
\therefore x = 1 &\text{ or } x = -3
\end{align*}
Solve for \(y\): \({y}^{3} = 3{y}^{2} + 16y + 12\)
\begin{align*}
\text{Let } a(y) &= {y}^{3}-3{y}^{2}-16y-12 \\
a(-1) &= (-1)^{3}-3(-1)^{2}-16(-1)-12 \\
&= -1-3+16-12 \\
&= 0 \\
\therefore a(y) &= (y +1)(y^{2} -4y -12) \\
&= (y+1)(y-6)(y+2) \\
\therefore 0 &= (y+1)(y-6)(y+2) \\
\therefore y = -1 &\text{ or } y=6 \text{ or } y =-2
\end{align*}
Solve for \(m\): \(m({m}^{2}-m-4) = - 4\)
\begin{align*}
\text{Let } a(m) &= {m}^{3}-{m}^{2}-4m+4 \\
a(1) &= (1)^{3}-(1)^{2}-4(1)+4 \\
&= 1 - 1 - 4 +4 \\
&= 0 \\
\therefore a(m) &= (m-1)(m^{2} - 4) \\
&= (m-1)(m+2)(m-2) \\
\therefore 0 &= (m-1)(m+2)(m-2) \\
\therefore m = 1 &\text{ or } m=2 \text{ or } m =-2
\end{align*}
Solve for \(x\): \({x}^{3}-{x}^{2}=3\left(3x+2\right)\)
\begin{align*}
{x}^{3}-{x}^{2} &= 3\left(3x+2\right) \\
{x}^{3}-{x}^{2} &= 9x + 6 \\
{x}^{3}-{x}^{2} -9x - 6 &= 0 \\
\text{Let } x =-2: \quad (-2)^{3}-(-2)^{2} -9(-2) - 6 \\
&= -8 -4 +18 - 6 \\
&= 0 \\
\therefore (x + 2) & \text{ is a factor} \\
(x + 2)(x^{2} - 3x -3) &= 0 \\
\text{Using quadratic formula to solve for } & x: x^{2} - 3x - 3 = 0 \\
a = 1; \quad b &= -3; \quad c=-3 \\
x &= \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\
&= \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)} \\
&= \frac{3 \pm \sqrt{9 + 12}}{2} \\
&= \frac{3 \pm \sqrt{21}}{2} \\
\therefore x = -2 &\text{ or } x =\frac{3 + \sqrt{21}}{2} \text{ or } x
=\frac{3 - \sqrt{21}}{2}
\end{align*}
Solve for \(x\) if \(2{x}^{3}-3{x}^{2}-8x=3\).
\begin{align*}
2{x}^{3}-3{x}^{2}-8x &= 3 \\
2{x}^{3}-3{x}^{2}-8x -3 &= 0 \\
\text{Let } a(x) &= 2{x}^{3}-3{x}^{2}-8x -3 \\
a(-1) &= 2(-1)^{3}-3(-1)^{2}-8(-1) -3 \\
&= -2 -3 + 8 -3 \\
&= 0 \\
\therefore a(x) &= (x + 1)(2x^{2} -5x -3) \\
&= (x + 1)(2x + 1)(x - 3) \\
\therefore 0 &= (x + 1)(2x + 1)(x - 3) \\
\therefore x = -1 &\text{ or } x = - \frac{1}{2} \text{ or } x= 3
\end{align*}
Solve for \(x\): \(16\left(x+1\right)={x}^{2}\left(x+1\right)\)
\begin{align*}
16\left(x+1\right) &= {x}^{2}\left(x+1\right) \\
16x + 16 &= {x}^{3} + {x}^{2} \\
0 &= {x}^{3} + {x}^{2} - 16x - 16 \\
\text{Let } a(x) &= {x}^{3} + {x}^{2} - 16x - 16 \\
a(-1) &= (-1)^{3} + (-1)^{2} - 16(-1) - 16 \\
&= -1 + 1 +16 -16 \\
&= 0 \\
\therefore a(x) &= (x + 1)(x^{2} - 16) \\
&= (x + 1)(x - 4)(x +4) \\
\therefore 0 &= (x + 1)(x - 4)(x +4) \\
\therefore x = -1 &\text{ or } x = 4 \text{ or } x= -4
\end{align*}
\(2{x}^{3}-{x}^{2}-2x+2=Q\left(x\right).\left(2x-1\right)+R\) for all values
of \(x\). What is the value of \(R\)?
\begin{align*}
\text{Let } a(x) &= 2{x}^{3}-{x}^{2}-2x+2 \\
R = a \left( \frac{1}{2} \right) &= 2\left( \frac{1}{2} \right)^{3}-\left(
\frac{1}{2} \right)^{2}-2\left( \frac{1}{2} \right) + 2 \\
&= 2 \left( \frac{1}{8} \right) - \left( \frac{1}{4} \right) - 1 + 2 \\
&= \frac{1}{4} - \frac{1}{4} +1 \\
&= 1 \\
\therefore R &= 1
\end{align*}
Use the factor theorem to solve the following equation for \(m\):
\(8{m}^{3}+7{m}^{2}-17m+2=0\)
\begin{align*}
\text{Let } a(m) &= 8{m}^{3}+7{m}^{2}-17m+2 \\
a(1) &= 8(1)^{3}+7(1)^{2}-17(1)+2 \\
&= 8 + 7 -17 + 2 \\
&= 0 \\
\therefore a(m) &= (m - 1)(8m^{2} + 15m - 2) \\
&= (m - 1)(8m - 1)(m + 2) \\
\therefore 0 &= (m - 1)(8m - 1)(m + 2) \\
\therefore m = 1 &\text{ or } m = \frac{1}{8} \text{ or } m = -2
\end{align*}
Hence, or otherwise, solve for \(x\):
\({2}^{3x+3}+7.{2}^{2x}+2=17.{2}^{x}\)
\begin{align*}
{2}^{3x+3}+7 \cdot {2}^{2x}+2 &= 17 \cdot {2}^{x} \\
{2}^{3x} \cdot 2^{3} +7 \cdot {2}^{2x}+2 &= 17 \cdot {2}^{x} \\
8 \cdot \left( {2}^{x} \right)^{3} +7 \cdot \left( {2}^{x} \right)^{2} -
17 \cdot {2}^{x} + 2 &= 0 \\
\text{which we can compare with } a(m) &= 8{m}^{3}+7{m}^{2}-17m+2 \\
\text{Let } 2^{x} &= m \\
\text{ and from part (a) we know that } m = 1 &\text{ or } m =
\frac{1}{8} \text{ or } m = -2 \\
\text{So } 2^{x} &= 1 \\
2^{x} &= 2^{0} \\
\therefore x &= 0 \\
\text{Or } 2^{x} &= \frac{1}{8} \\
2^{x} &= 2^{-3} \\
\therefore x &= -3 \\
\text{Or } 2^{x} &= -2 \\
\therefore & \text{ no solution}
\end{align*}
Find the value of \(R\) if \(x-1\) is a factor of \(h(x)= (x - 6) \cdot Q(x)
+ R\) and \(Q(x)\) divided by \(x-1\) gives a remainder of \(\text{8}\).
\begin{align*}
h(x) &= (x - 6) \cdot Q(x) + R \\
h(1) &= (1 - 6) \cdot Q(1) + R \\
\therefore 0 &= -5 \cdot Q(1) + R \\
\text{And } Q(1) &= 8 \\
0 &= -5(8) + R \\
\therefore R &= 40
\end{align*}
Determine the values of \(p\) for which the function
\[f\left(x\right)=3{p}^{3}-\left(3p-7\right){x}^{2}+5x-3\]
leaves a remainder of \(\text{9}\) when it is divided by
\(\left(x-p\right)\).
\begin{align*}
f\left(x\right) &= 3{p}^{3}-\left(3p-7\right){x}^{2}+5x-3 \\
\therefore f(p) &= 3{p}^{3}-\left(3p-7\right){p}^{2}+5p-3 \\
&= 3{p}^{3}- 3p^{3} + 7p^{2} + 5p - 3 \\
&= 7p^{2} + 5p - 3 \\
f(p) &= 9 \\
\therefore 9 &= 7p^{2} + 5p - 3 \\
0 &= 7p^{2} + 5p - 12 \\
0 &= (7p + 12)(p - 1) \\
\therefore p = -\frac{12}{7} &\text{ or } p = 1
\end{align*}
Alternative (long) method:
We first take out the factor using long division:
\begin{align*}
&\qquad \quad \underline{(7-3p)x + (5 + 7p -3p^{2})} \\
&(x-p) | (7-3p)x^{2} + 5x + (3p^{3} - 3)\\
&\quad \quad - \underline{\lbrace (7-3p)x^{2} - p(7-3p)x \rbrace} \\
&\qquad \qquad \qquad \qquad 0 + 5x + p(7-3p)x + (3p^{3}-3)\\
&\qquad \qquad \qquad \qquad \qquad 5x + 7px -3p^{2}x + 3p^{3}x \\
&\qquad \qquad \qquad \qquad \qquad [5 + 7p -3p^{2}]x + 3p^{3} - 3
\\
&\qquad \qquad \qquad \qquad \quad - \underline{\lbrace [5 + 7p
-3p^{2}]x -p(5 + 7p - 3p^{2}) \rbrace}\\
&\qquad \qquad \qquad \qquad \qquad 0 + 3p^{3} - 3 + 5p + 7p^{2}
-3p^{3}
\end{align*}
We take the remainder and set it equal to 9:
\begin{align*}
-3 + 5p + 7p^{2} & = 9 \\
7p^{2} + 5p - 12 & = 0\\
(7p+12)(p-1) & = 0\\
\therefore p = -\frac{12}{7} & \text{ or } p = 1
\end{align*}
Calculate \(t\) and \(Q(x)\) if \(x^{2} + tx + 3 = (x + 4) \cdot Q(x) - 17\).
\begin{align*}
x^{2} + tx + 20 &= (x + 4) \cdot Q(x) \\
\text{Let } f(x) &= x^{2} + tx + 20 \\
f(-4) &= (-4)^{2} + t(-4) + 20 \\
0 &= 16 -4t + 20 \\
4t &= 36 \\
\therefore t &= 9 \\
& \\
x^{2} + 9x + 20 &= (x + 4) \cdot Q(x) \\
(x + 4)(x + 5)&= (x + 4) \cdot Q(x) \\
\therefore Q(x) &= x + 5
\end{align*}
