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# End of chapter exercises

## End of chapter exercises

Textbook Exercise 5.7

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*}

Show that $$x-2$$ is a factor of $$3{x}^{3}-11{x}^{2}+12x-4$$.

\begin{align*} \text{Let } a(x) &= 3{x}^{3}-11{x}^{2}+12x-4 \\ a(2) &= 3(2)^{3}-11(2)^{2}+12(2)-4 \\ &= 24 - 44 +24 - 4 \\ &= 0 \\ \therefore (x-2) &\text{ is a factor of } a(x) \end{align*}

Hence, by factorising completely, solve the equation:

$$3{x}^{3}-11{x}^{2}+12x-4=0$$
\begin{align*} 3{x}^{3}-11{x}^{2}+12x-4 &= 0 \\ (x -2)(3x^{2} - 5x + 2) &= 0 \\ \therefore (x - 2)(3x - 2)(x - 1) &= 0 \\ \therefore x = 2 &\text{ or } x = \frac{2}{3} \text{ or } x = 1 \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:

Calculate $$t$$ and $$Q(x)$$ if $$x^{2} + tx + 3 = (x + 4) \cdot Q(x) - 17$$.