We think you are located in South Africa. Is this correct?

# Practise now to improve your marks

You can do it! Let us help you to study smarter to achieve your goals. Siyavula Practice guides you at your own pace when you do questions online.

## End of chapter exercises

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