The Factor Theorem (Edexcel IGCSE Maths)

Factor Theorem

Spec point — 3D

What is the factor theorem?

• The factor theorem is used to find the linear factors of a function

  • This is closely related to finding the roots (or solutions) of a function or equation

• For a function $\mathrm{f} ( x )$, the factor theorem tells us that

  • If $\mathrm{f} ( a ) = 0$, then $( x - a )$ is a factor of $\mathrm{f} ( x )$
  • If $( x - a )$ is a factor of $\mathrm{f} ( x )$, then $\mathrm{f} ( a ) = 0$

How do I use the factor theorem?

• Consider the function $\mathrm{f} ( x )$ where $( x - a )$ is a factor

  • Then by the factor theorem we know that $\mathrm{f} ( a ) = 0$

    • I.e., $x=a$ is a solution to the equation $\mathrm{f} ( x ) = 0$

• Or consider the function $\mathrm{f} ( x )$ where $\mathrm{f} ( a ) = 0$

  • Then by the factor theorem we know that $( x - a )$ is a factor of $\mathrm{f} ( x )$

  • Therefore $f(x)=(x-a)\times Q(x)$

    • where $Q ( x )$ is a function that is also a factor of $\mathrm{f} ( x )$

  • Hence $\frac{ \mathrm{f} ( x )}{( x - a )} = Q ( x )$

    • I.e. $Q ( x )$ is the quotient when $\mathrm{f} ( x )$ is divided by $( x - a )$
    • And the remainder is equal to zero

• If the linear factor has a coefficient of x *(other than 1) you must first factorise out the coefficient

  • For the linear factor $( b x - c ) = b ( x - \frac{c}{b} )$

    • $\mathrm{f} ( \frac{c}{b} ) = 0$
    • $\mathrm{f} ( x ) = b ( x - \frac{c}{b} ) \times Q ( x ) = ( b x - c ) \times Q ( x )$

Exam Hint Be careful with the minus sign in a factor $( x - a )$.

• That means $a$ is a solution to $f ( x ) = 0$, not $-a$ !

If you are looking for integer solutions to $\mathrm{f} ( x ) = 0$ (where $\mathrm{f} ( x )$ is a polynomial)

• those solutions will always be factors of the constant term in $\mathrm{f} ( x )$

Worked Example (a) Consider the function$f(x)=x^{3}-2x^{2}-x+2$. Given that $x=2$ is a solution to the equation $\mathrm{f} ( x ) = 0$, write down a linear factor of $\mathrm{f} ( x )$.

Answer:

By the factor theorem, if $\mathrm{f} ( a ) = 0$ then $( x - a )$ is a factor of $\mathrm{f} ( x )$

$x-2$* *

(b) Use the factor theorem to determine whether $( x + 1 )$ is a factor of $g(x)=2x^{3}+3x^{2}-x+5$.

Answer:

By the factor theorem, $( x - a )$ can only be a factor of $\mathrm{g} ( x )$ if $\mathrm{g} ( a ) = 0$

But be careful, here $a$ is equal to $-1$, not $1$

$\begin{gathered} \mathrm{g} ( - 1 ) & = 2 ( - 1 )^{3} + 3 ( - 1 )^{2} - ( - 1 ) + 5 \\ = - 2 + 3 + 1 + 5 \\ = 7 \end{gathered}$

Final answer: $g(-1)\neq0$, so$(x+1)$is not a factor of$g(x)$

(c) Given that $( 2 x - 3 )$ is a factor of $\mathrm{h} ( x ) = 2 x^{3} - b x^{2} + 7 x - 6$, find the value of $b$.

Answer:

$( 2 x - 3 ) = 2 ( x - \frac{3}{2} )$, so $( x - \frac{3}{2} )$ is a factor of $\mathrm{h} ( x )$

Therefore by the factor theorem, $\mathrm{h} ( \frac{3}{2} ) = 0$

$\begin{gathered} 2 ( \frac{3}{2} )^{3} - b ( \frac{3}{2} )^{2} + 7 ( \frac{3}{2} ) - 6 & = 0 \\ \frac{27}{4} - \frac{9}{4} b + \frac{21}{2} - 6 & = 0 \\ \frac{45}{4} - \frac{9}{4} b & = 0 \\ \frac{9}{4} b & = \frac{45}{4} \\ b = \frac{4}{9} \times \frac{45}{4} \end{gathered}$

$b=5$