Challenge · Factor Theorem

20 marks

These need more than one step. The factor theorem says $(x - a)$ is a factor of $f(x)$ exactly when $f(a) = 0$ – note the sign: to test the factor $(x + 2)$ you substitute $x = -2$. Once you have one factor, divide it out and factorise what is left. There is a worked example to show the shape of the reasoning; hints are at the foot of the sheet. No calculator.

Worked example: how to start

Show that $(x - 2)$ is a factor of $f(x) = x^{3} - 3x^{2} + 4$.

The factor $(x - 2)$ corresponds to testing $x = 2$.

Substitute: $f(2) = 2^{3} - 3 \times 2^{2} + 4 = 8 - 12 + 4$.

This equals $0$, so by the factor theorem $(x - 2)$ is a factor.

1A student claims that $(x + 1)$ is a factor of $f(x) = x^{3} + 3x^{2} - 4$. Is the student correct? Justify your answer using the factor theorem.[2 marks]

Answer:

2It is given that $f(x) = x^{3} - x^{2} - 8x - 4$ has $(x + 2)$ as a factor. Factorise $f(x)$ as far as possible, and hence write down the exact solutions of $f(x) = 0$, giving any non-integer solutions in surd form.[4 marks]

Answer:

3It is given that $f(x) = x^{3} - 2x^{2} + kx + 5$ leaves a remainder of $11$ when divided by $(x - 3)$. Work out the value of $k$.[4 marks]

Answer:

4Solve the equation $x^{3} - 7x + 6 = 0$.[5 marks]

Answer:

5It is given that $f(x) = x^{3} + ax^{2} + bx - 10$, that $(x - 1)$ is a factor, and that $f(2) = 0$. Work out the values of $a$ and $b$.[5 marks]

Answer:

★Extension: Explain why, if a cubic with whole-number coefficients and constant term $-12$ has a whole-number root, that root must be a factor of $12$.

Stuck? Hints (don't peek unless you need to)1. Substitute the value that would make the bracket zero and check whether the result is zero.2. Divide out the given factor first. If the quadratic left over will not factorise with whole numbers, that is your signal to use the quadratic formula instead.3. The remainder theorem says the remainder when dividing by $(x - 3)$ equals $f(3)$. Substitute $x = 3$ and set the result equal to $11$.4. Try small whole numbers that divide the constant $6$ until one gives zero. Then divide out that factor and solve the quadratic.5. Each condition gives an equation in $a$ and $b$. Form both, then solve them as a pair of simultaneous equations.

Solutions & mark scheme · Factor Theorem

Total: 20 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1A student claims that $(x + 1)$ is a factor of $f(x) = x^{3} + 3x^{2} - 4$. Is the student correct? Justify your answer using the factor theorem.[2]

Model solution

The bracket $(x + 1)$ is zero when $x = -1$, so substitute $x = -1$ into $f(x)$.

$f(-1) = (-1)^{3} + 3(-1)^{2} - 4 = -1 + 3 - 4 = -2$.

Since $f(-1) = -2$ and this is not 0, the factor theorem shows $(x+1)$ is not a factor, so the student is wrong.

Answer: No: $f(-1) = -2$, which is not 0, so $(x+1)$ is not a factor.

Marks

✔1substitutes $x = -1$ into $f(x)$ correctly

✔1evaluates $f(-1) = -2$, states it is not 0, and concludes the student is wrong

Model solution

Dividing $f(x)$ by $(x+2)$ gives $x^{2} - 3x - 2$, so $f(x) = (x+2)(x^{2}-3x-2)$.

Trying to factorise $x^{2}-3x-2$ with whole numbers fails: no pair of numbers multiplies to $-2$ and adds to $-3$, so the quadratic formula is needed instead.

$x = \dfrac{3 \pm \sqrt{9+8}}{2} = \dfrac{3 \pm \sqrt{17}}{2}$.

The three exact solutions are $x=-2$, $x=\dfrac{3+\sqrt{17}}{2}$ and $x=\dfrac{3-\sqrt{17}}{2}$.

Answer: $f(x) = (x+2)(x^{2}-3x-2)$; solutions $x=-2$, $x=\dfrac{3+\sqrt{17}}{2}$, $x=\dfrac{3-\sqrt{17}}{2}$

Marks

✔1divides by $(x+2)$ to get quotient $x^{2}-3x-2$

✔1recognises the quadratic does not factorise with whole numbers and applies the quadratic formula

✔1correct discriminant and surd, $x=\dfrac{3\pm\sqrt{17}}{2}$

✔1states all three exact solutions, $x=-2,\ \dfrac{3+\sqrt{17}}{2},\ \dfrac{3-\sqrt{17}}{2}$

3It is given that $f(x) = x^{3} - 2x^{2} + kx + 5$ leaves a remainder of $11$ when divided by $(x - 3)$. Work out the value of $k$.[4]

Model solution

By the remainder theorem, dividing by $(x - 3)$ gives a remainder equal to $f(3)$, so $f(3) = 11$.

$f(3) = 3^{3} - 2 \times 3^{2} + 3k + 5 = 27 - 18 + 3k + 5 = 14 + 3k$.

Setting this equal to the given remainder: $14 + 3k = 11$.

Solving: $3k = -3$, so $k = -1$.

Answer: $k = -1$

Marks

✔1uses the remainder theorem, $f(3) = 11$

✔1substitutes correctly, $f(3) = 14 + 3k$

✔1forms the equation $14 + 3k = 11$

✔1$k = -1$

4Solve the equation $x^{3} - 7x + 6 = 0$.[5]

Model solution

Try $x = 1$: $1 - 7 + 6 = 0$, so $(x - 1)$ is a factor.

Dividing $x^{3} - 7x + 6$ by $(x - 1)$ gives $x^{2} + x - 6$.

Factorising: $x^{2} + x - 6 = (x + 3)(x - 2)$.

So the equation is $(x - 1)(x + 3)(x - 2) = 0$.

The solutions are $x = 1$, $x = -3$ and $x = 2$.

Answer: $x = 1$, $x = 2$ and $x = -3$

Marks

✔1finds $f(1) = 0$

✔1divides by $(x-1)$

✔1quotient $= x^{2} + x - 6$

✔1factorises to $(x+3)(x-2)$

✔1all three solutions

5It is given that $f(x) = x^{3} + ax^{2} + bx - 10$, that $(x - 1)$ is a factor, and that $f(2) = 0$. Work out the values of $a$ and $b$.[5]

Model solution

$(x-1)$ a factor means $f(1) = 0$: $1 + a + b - 10 = 0$, so $a + b = 9$.

$f(2) = 0$ gives $8 + 4a + 2b - 10 = 0$, so $4a + 2b = 2$, that is $2a + b = 1$.

Subtracting the first from the second: $(2a + b) - (a + b) = 1 - 9$, so $a = -8$.

Then $b = 9 - a = 9 - (-8) = 17$.

So $a = -8$ and $b = 17$.

Answer: $a = -8$ and $b = 17$

Marks

✔1uses $f(1) = 0$

✔1$a + b = 9$

✔1uses $f(2) = 0$

✔1$2a + b = 1$

✔1$a = -8$, $b = 17$

★Extension

Suppose the cubic factorises as $(x - r)(x^{2} + px + q)$ with $r$, $p$ and $q$ whole numbers.

Multiplying out, the constant term of the product is $-rq$.

So $-rq = -12$, which means $rq = 12$ and therefore $r$ divides $12$ exactly.

That is why testing only the factors of the constant term – here $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$ – is enough to find any whole-number root.