A-Level Maths | Cubics
Divide the cubic by the linear factor, and fully factorise the cubic if possible.
(a) $x^3+2x^2-x-2$ by $x+1$
(b) $x^3+2x^2-9x+2$ by $x-2$
(c) $20+x+3x^2+x^3$ by $x+4$
(d) $2x^3-x^2-4x+3$ by $x-1$
(e) $6x^3-19x^2-73x+90$ by $x-5$
(f) $-x^3+5x^2+10x-8$ by $x+2$
(g) $x^3-2x+21$ by $x+3$
(h) $3x^3+16x^2+72$ by $x+6$
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(a) Quotient $x^2+x-2$; fully factorised: $(x-1)(x+1)(x+2)$
(b) Quotient $x^2+4x-1$ (does not factorise further over rationals)
(c) Quotient $x^2-x+5$ (does not factorise further — no real roots)
(d) Quotient $2x^2+x-3$; fully factorised: $(x-1)^2(2x+3)$
(e) Quotient $6x^2+11x-18$ (does not factorise further over rationals)
(f) Quotient $-x^2+7x-4$ (does not factorise further over rationals)
(g) Quotient $x^2-3x+7$ (does not factorise further — no real roots)
(h) Quotient $3x^2-2x+12$ (does not factorise further — no real roots)
Use the factor theorem to determine whether or not
(a) $(x-1)$ is a factor of $x^3+2x^2-2x-1$
(b) $(x+2)$ is a factor of $x^3-5x^2-9x+2$
(c) $(x-3)$ is a factor of $x^3-x^2-14x+27$
(d) $(x+6)$ is a factor of $2x^3+13x^2+2x-24$
(e) $(2x+1)$ is a factor of $2x^3-5x^2+7x-14$
(f) $(3x-2)$ is a factor of $2-17x+25x^2-6x^3$
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(a) $f(1)=0$ — yes, a factor
(b) $f(-2)=-8$ — no, not a factor
(c) $f(3)=3$ — no, not a factor
(d) $f(-6)=0$ — yes, a factor
(e) $f(-\frac{1}{2})=-19$ — no, not a factor
(f) $f(\frac{2}{3})=0$ — yes, a factor
Use the remainder theorem to find the remainder obtained in dividing
(a) $x^3+4x^2-x+6$ by $x-2$
(b) $x^3-2x^2+7x+1$ by $x+1$
(c) $2x^3+x^2-9x+17$ by $x+5$
(d) $8x^3+4x^2-6x-3$ by $2x-1$
(e) $2x^3-3x^2-20x-7$ by $2x+1$
(f) $3x^3-6x^2+2x-7$ by $3x-2$
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(a) $28$
(b) $-9$
(c) $-163$
(d) $-4$
(e) $2$
(f) $-\frac{67}{9}$
$$f(x) \equiv x^3-2x^2-11x+12$$
(a) Show that $(x-1)$ is a factor of $f(x)$.
(b) Hence, express $f(x)$ as the product of three linear factors.
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(a) $f(1) = 1-2-11+12 = 0$, so $(x-1)$ is a factor.
(b) Dividing $f(x)$ by $(x-1)$ gives $x^2-x-12$, which factorises as $(x-4)(x+3)$.
$$f(x) = (x-1)(x-4)(x+3)$$
Given that $x=-2$ is a solution to the equation $$g(x)\equiv x^3+7x^2+7x-6=0$$
(a) express $g(x)$ as the product of a linear factor and a quadratic factor,
(b) find, to 2 decimal places, the other two solutions to the equation $g(x)=0$.
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(a) Dividing $g(x)$ by $(x+2)$ gives $x^2+5x-3$.
$$g(x) = (x+2)(x^2+5x-3)$$
(b) Solving $x^2+5x-3=0$ by the quadratic formula:
$$x = \frac{-5\pm\sqrt{25+12}}{2} = \frac{-5\pm\sqrt{37}}{2}$$
$$x \approx 0.54 \text{ or } x \approx -5.54$$
By first finding a linear factor, fully factorise $$x^3-2x^2-5x+6$$
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Testing $x=1$: $1-2-5+6=0$, so $(x-1)$ is a factor.
Dividing by $(x-1)$ gives $x^2-x-6 = (x-3)(x+2)$.
$$x^3-2x^2-5x+6 = (x-1)(x-3)(x+2)$$
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