5.5 Solving cubic equations | Polynomials

5.5 Solving cubic equations (EMCGX)

Now that we know how to factorise cubic polynomials, it is also easy to solve cubic equations of the form \(a{x}^{3}+b{x}^{2}+cx+d=0\).

Worked example 13: Solving cubic equations

Solve: \(6{x}^{3}-5{x}^{2}-17x+6 = 0\)

Find one factor using the factor theorem

Let \(f(x) = 6{x}^{3}-5{x}^{2}-17x+6\)

Try

\(f\left(1\right)=6{\left(1\right)}^{3}-5{\left(1\right)}^{2}-17\left(1\right)+6=6-5-17+6=-10\)

Therefore \(\left(x-1\right)\) is not a factor.

Try

\(f\left(2\right)=6{\left(2\right)}^{3}-5{\left(2\right)}^{2}-17\left(2\right)+6=48-20-34+6=0\)

Therefore \(\left(x-2\right)\) is a factor.

Factorise by inspection

\[6{x}^{3}-5{x}^{2}-17x+6=\left(x-2\right)\left(6{x}^{2}+7x-3\right)\]

Factorise fully

\[6{x}^{3}-5{x}^{2}-17x+6 = \left(x-2\right) \left(2x+3\right) \left(3x-1\right)\]

Solve the equation

\begin{align*} 6{x}^{3}-5{x}^{2}-17x+6 & = 0 \\ \left(x-2\right)\left(2x+3\right)\left(3x-1\right) & = 0 \\ x = 2 \text{ or } x & = \frac{1}{3} \text{ or } x = -\frac{3}{2} \end{align*}

Sometimes it is not possible to factorise a quadratic expression using inspection, in which case we use the quadratic formula to fully factorise and solve the cubic equation.

\[x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\]

Worked example 14: Solving cubic equations

Solve for \(x\): \(0 = {x}^{3}-2{x}^{2}-6x+4\)

Use the factor theorem to determine a factor

Let \(f(x)= {x}^{3}-2{x}^{2}-6x+4\)

Try

\(f\left(1\right)={\left(1\right)}^{3}-2{\left(1\right)}^{2}-6\left(1\right)+4=1-2-6+4=-3\)

Therefore \(\left(x-1\right)\) is not a factor.

Try

\(f\left(2\right)={\left(2\right)}^{3}-2{\left(2\right)}^{2}-6\left(2\right)+4=8-8-12+4=-8\)

Therefore \(\left(x-2\right)\) is not a factor.

\(f\left(-2\right)={\left(-2\right)}^{3}-2{\left(-2\right)}^{2}-6\left(-2\right)+4=-8-8+12+4=0\)

Therefore \(\left(x+2\right)\) is a factor.

Factorise by inspection

\[{x}^{3}-2{x}^{2}-6x+4 = \left(x+2\right) \left({x}^{2}-4x+2\right)\]

\({x}^{2}-4x+2\) cannot be factorised any further and we are left with

\(\left(x+2\right)\left({x}^{2}-4x+2\right)=0\)

Solve the equation

\begin{align*} \left(x+2\right)\left({x}^{2}-4x+2\right) & = 0 \\ \left(x+2\right)=0 & \text{ or } \left({x}^{2}-4x+2\right)=0 \end{align*}

Apply the quadratic formula for the second bracket

Always write down the formula first and then substitute the values of \(a,b\) and \(c\).

\[a = 1; \qquad b = -4; \qquad c = 2\]

\begin{align*} x & = \frac{-b±\sqrt{{b}^{2}-4ac}}{2a} \\ & = \frac{-\left(-4\right)±\sqrt{{\left(-4\right)}^{2} -4\left(1\right)\left(2\right)}}{2\left(1\right)} \\ & = \frac{4±\sqrt{8}}{2} \\ & = 2±\sqrt{2} \end{align*}

Final solutions

\(x=-2\) or \(x=2±\sqrt{2}\)

temp text

Solving cubic equations

Textbook Exercise 5.6

\(x^{3} + x^{2} - 16x = 16\)

\begin{align*} x^{3} + x^{2} - 16x &= 16 \\ x^{3} + x^{2} - 16x - 16 &= 0 \\ \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*}

\(-n^{3} - n^{2} + 22n + 40 = 0\)

\begin{align*} n^{3} + n^{2} - 22n - 40 &= 0 \\ \text{Let } a(n) &= n^{3} + n^{2} - 22n - 40 \\ a(-2) &= (-2)^{3} + (-2)^{2} - 22(-2) - 40 \\ &= -8 +4 +44 -40\\ &= 0 \\ \therefore a(n) &= (n + 2)(n^{2} -n -20) \\ &= (n+2)(n-5)(n+4) \\ \therefore 0 &= (n+2)(n-5)(n+4) \\ \therefore n=-2 & \text{ or } n = -4 \text{ or } n = 5 \end{align*}

\(y(y^{2} + 2y) = 19y + 20\)

\begin{align*} y(y^{2} + 2y) &= 19y + 20 \\ y^{3} + 2y^{2} - 19y - 20 &= 0\\ \text{Let } a(y) &= y^{3} + 2y^{2} - 19y - 20 \\ a(-1) &= (-1)^{3} + 2(-1)^{2} - 19(-1) - 20 \\ &= -1 +2 +19 -20 \\ &= 0 \\ \therefore a(y) &= (y + 1)(y^{2} +y -20) \\ &= (y + 1)(y + 5)(y - 4) \\ \therefore 0 &= (y + 1)(y + 5)(y - 4) \\ \therefore y=-1 & \text{ or } y = 4 \text{ or } y = -5 \end{align*}

\(k^{3} + 9k^{2} + 26k + 24 = 0\)

\begin{align*} \text{Let } a(k) &= k^{3} + 9k^{2} + 26k + 24 \\ a(-2) &= (-2)^{3} + 9(-2)^{2} + 26(-2) + 24 \\ &= -8 + 36 - 52 +24 \\ &= 0 \\ \therefore a(k) &= (k + 2)(k^{2} + 7k + 12 ) \\ &= (k + 2)(k+3)(k + 4) \\ \therefore 0 &= (k + 2)(k+3)(k + 4) \\ \therefore k = -2 & \text{ or } k = -3 \text{ or } k = -4 \end{align*}

\(x^{3} + 2x^{2} - 50 = 25x\)

\begin{align*} x^{3} + 2x^{2} - 50 &= 25x \\ x^{3} + 2x^{2} - 25x - 50 &= 0 \\ \text{Let } a(x) &= x^{3} + 2x^{2} - 25x - 50 \\ a(-2) &= (-2)^{3} + 2(-2)^{2} - 25(-2) - 50 \\ &= -8 + 8 + 50 -50 \\ &= 0 \\ \therefore a(x) &= (x + 2)(x^{2} - 25) \\ &= (x + 2)(x-5)(x + 5) \\ \therefore 0 &= (x + 2)(x-5)(x + 5) \\ \therefore x = -2 & \text{ or } x = 5 \text{ or } x = -5 \end{align*}

\(-p^{3} + 19p = 30\)

\begin{align*} -p^{3} + 19p -30 & = 0 \\ p^{3} - 19p + 30 & = 0 \\ \text{Let } a(p) &= p^{3} - 19p + 30 \\ a(3) &= (3)^{3} - 19(3) + 30 \\ &= 27 -57 + 30 \\ &= 0 \\ \therefore a(p) &= (p - 3)(p^{2} + 3p - 10) \\ &= (p-3)(p - 2)(p + 5) \\ \therefore 0 &= (p-3)(p - 2)(p + 5) \\ \therefore p = 3 & \text{ or } p = 2 \text{ or } p = -5 \end{align*}

\(6x^{2} - x^{3} = 5x + 12\)

\begin{align*} 0 = x^{3} - 6x^{2} + 5x + 12 \\ \text{Let } a(x) &= x^{3} - 6x^{2} + 5x + 12 \\ a(-1) &= (-1)^{3} - 6(-1)^{2} + 5(-1) + 12 \\ &= -1 -6 -5 + 12 \\ &= 0 \\ \therefore a(x) &= (x + 1)(x^{2} - 7x + 12) \\ &= (x + 1)(x-3)(x - 4) \\ \therefore 0 &= (x + 1)(x-3)(x - 4) \\ \therefore x = -1 & \text{ or } x = 3 \text{ or } x = 4 \end{align*}

5.4 Factor theorem

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5.6 Summary