It is given that $(x - 1)$ is a factor of $f(x) = x^{3} + kx - 5$. Work out $k$.
1
$(x-1)$ a factor means $f(1) = 0$
2
$f(1) = 1 + k - 5$, and setting it to zero gives $k - 4 = 0$
3
so $k = -4$
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③Solution 3
Factorise $f(x) = x^{3} - x^{2} - 4x + 4$ completely, given that $(x - 1)$ is a factor.
1
dividing $f(x)$ by $(x-1)$ gives $x^{2} - 4$
2
$x^{2} - 4$ will not factorise because there is no middle term
3
so $f(x) = (x - 1)\left(x^{2} - 4\right)$
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④Solution 4
Solve $x^{3} - 4x^{2} + x + 6 = 0$, given that $x = -1$ is one solution.
1
since $x = -1$ is a solution, $(x + 1)$ is a factor
2
dividing gives $x^{2} - 5x + 6$, which factorises as $(x-2)(x-3)$
3
so the solutions are $x = -1$, $x = -2$ and $x = -3$
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Answers · Factor Theorem
Spot the mistake
① Solution 1 line 1: the bracket $(x+2)$ is zero when $x = -2$, not $x = 2$ – the sign always flips when you go from the factor to the value you test
Fix: substitute $x = -2$ $f(-2) = 0,\ \text{so it IS a factor}$
② Solution 2 line 3: $k - 4 = 0$ gives $k = 4$, not $-4$ – the sign has been flipped when rearranging
Fix: $k - 4 = 0$, so $k = 4$ $4$
③ Solution 3 line 2: $x^{2} - 4$ is a difference of two squares, so it does factorise: $(x+2)(x-2)$. A missing middle term is exactly the signal for that
Fix: $x^{2} - 4 = (x+2)(x-2)$ $(x-1)(x+2)(x-2)$
④ Solution 4 line 3: the factor $(x - 2)$ gives the root $x = +2$, not $-2$ – the sign flips going from a bracket to its root, and here it has been left unchanged
Fix: the roots are $x = -1$, $x = 2$ and $x = 3$ $x = -1,\ 2,\ 3$
Factor Theorem