The factor theoremAQA A-Level Maths: Revision notes

For a polynomial $f(x)$, $(x-a)$ is a factor of $f(x)$ if and only if $f(a)=0$. In words: if substituting $x=a$ gives zero then $(x-a)$ is a factor, and if $(x-a)$ is a factor then $f(a)=0$. This works because if $f(x)=(x-a)q(x)$ then $f(a)=0\times q(a)=0$. For a factor of the form $(ax-b)$, the value to test is $x=\frac ba$, the solution of $ax-b=0$. So $(2x-3)$ is a factor of $f(x)$ when $f\left(\frac32\right)=0$, and $(3x+2)$ is a factor when $f\left(-\frac23\right)=0$. Note the sign: $(x+3)$ needs $f(-3)=0$.

Substitute values into $f(x)$ and look for $0$. For $f(x)=2x^3-x^2-7x+6$: $f(1)=2-1-7+6=0$, so $(x-1)$ is a factor; $f(-1)=-2-1+7+6=10\ne0$, so $(x+1)$ is not. Which values to try? A whole-number root must divide the constant term (here 6), so try $\pm1,\pm2,\pm3,\pm6$. If the leading coefficient is not 1, a fractional root has a numerator which divides the constant term and a denominator which divides the leading coefficient, such as $\pm\frac32$ or $\pm\frac12$. Start with the small values $1$, $-1$, $2$, $-2$. If $f(a)\ne0$ then $(x-a)$ is definitely not a factor, so show the calculation and conclude clearly.

To factorise a cubic completely:

1. Find one factor $(x-a)$ using $f(a)=0$ (or use the factor you are given).
2. Divide by it, by algebraic division or by comparing coefficients, to get a quadratic.
3. Factorise the quadratic. Example: $f(x)=2x^3-x^2-7x+6$ with $(x-1)$ a factor gives $f(x)=(x-1)(2x^2+x-6)=(x-1)(2x-3)(x+2)$. For a quartic, find two factors, multiply them to make a quadratic, and divide by that. Example: $x^4-3x^3-7x^2+15x+18$ has factors $(x+1)$ and $(x-3)$, so divide by $x^2-2x-3$ to get $x^2-x-6=(x-3)(x+2)$, and $f(x)=(x+1)(x-3)^2(x+2)$.

Once $f(x)$ is fully factorised, each linear factor gives a solution of $f(x)=0$, because a product is zero only if one factor is zero. Example: $p(x)=x^3-3x^2-4x+12$. Then $p(x)=x^2(x-3)-4(x-3)=(x-3)(x-2)(x+2)$, so $p(x)=0$ gives $x=3$, $x=2$ or $x=-2$. A repeated factor gives a repeated solution: $(x-2)^2(x+3)=0$ gives $x=2$ (repeated) or $x=-3$. Always state the solutions at the end, and when a factor is $(2x+1)$ the solution is $x=-\frac12$, not $-1$ or $\frac12$.

If a polynomial contains unknown constants and you know factors, substitute the corresponding values to form equations. Example: $g(x)=2x^3+ax^2+bx-12$ has factors $(x-2)$ and $(x+3)$. Then $g(2)=16+4a+2b-12=0$, so $2a+b=-2$; and $g(-3)=-54+9a-3b-12=0$, so $3a-b=22$. Adding the two equations gives $5a=20$, so $a=4$ and $b=-10$. Then $g(x)=2x^3+4x^2-10x-12=2(x-2)(x+3)(x+1)$. With one unknown, only one equation is needed, e.g. $p(x)=x^3+kx^2-4x+12$ with factor $(x-2)$: $8+4k-8+12=0$ gives $k=-3$. Check your values by substituting back into $f$.