Home Practice
For learners and parents For teachers and schools
Textbooks
Mathematics
Junior Secondary School 1 Junior Secondary School 2 Junior Secondary School 3
Full catalogue
Leaderboards
Learners Leaderboard Classes Leaderboard Schools Leaderboard
Pricing Support
Help centre Contact us
Log in

We think you are located in Nigeria. Is this correct?

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:

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*}