Factor & Remainder Theorems (Edexcel IGCSE Further Maths)

Factor Theorem

Spec point — 3B

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

    • 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 $( b x - c ) = b ( x - \frac{c}{b} )$

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

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 $\mathrm{f} ( x )$.

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 $g(x)=2x^{3}+3x^{2}-x+5$.

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 $1$

Final answer: $\begin{gathered} \mathrm{g} ( - 1 ) & = 2 ( - 1 )^{3} + 3 ( - 1 )^{2} - ( - 1 ) + 5 \\ = - 2 + 3 + 1 + 5 \\ = 7 \end{gathered}$ $g(-1)\neq0$, so$(x+1)$is not a factor of$g(x)$

c) It is 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 $b$.

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

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

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

$b=5$

Remainder Theorem

Spec point — 3B

What is the remainder theorem?

• The remainder theorem is used to find the remainder when we divide a polynomial function by a linear function

• When a polynomial function $\mathrm{f} ( x )$ is divided by a linear function $( x - a )$, the value of the remainder$R$ is given by $\mathrm{f} ( a ) = R$

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

How do I use the remainder theorem?

• Consider a polynomial function $\mathrm{f} ( x )$ and a linear function $( x - a )$

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

    • $Q ( x )$ is the quotient (also a polynomial function)
    • $R$ is the remainder(a real number)

  • This may also be written as $\mathrm{f} ( x ) = Q ( x ) \times ( x - a ) + R$

  • The remainder theorem tells us that $R = \mathrm{f} ( a )$

    • I.e. we don't need to do the algebraic division to find the remainder!

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

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

    • $R = \mathrm{f} ( \frac{c}{b} )$

Exam Hint

• Be careful with the minus sign in $( x - a )$

  • You need to put $a$ into $\mathrm{f} ( x )$ to find the remainder, not $-a$!

Worked Example a) Find the remainder when the function $\mathrm{f} ( x ) = 2 x^{4} - 2 x^{3} - x^{2} - 3 x + 1$ is divided by $( x - 2 )$.

We're dividing by $( x - a ) = ( x - 2 )$

So $a=2$

By the remainder theorem the remainder will be

$\begin{gathered} \mathrm{f} ( 2 ) & = 2 ( 2 )^{4} - 2 ( 2 )^{3} - ( 2 )^{2} - 3 ( 2 ) + 1 \\ = 32 - 16 - 4 - 6 + 1 \\ = 7 \end{gathered}$

Final answer: Remainder = 7

b) The remainder when $\mathrm{g} ( x ) = 2 x^{3} + x^{2} + b x + 1$ is divided by $( 2 x + 1 )$ is 3. Find the value of $b$.

$( 2 x + 1 ) = 2 ( x + 1 \frac{1}{2} ) = 2 ( x - ( - 1 \frac{1}{2} ) )$

So here the value of $a$ to use is $-\frac{1}{2}$

By the remainder theorem the remainder will be equal to $\mathrm{g} ( - 1 \frac{1}{2} )$

$\begin{gathered} 2 ( - 1 \frac{1}{2} )^{3} + ( - 1 \frac{1}{2} )^{2} + b ( - 1 \frac{1}{2} ) + 1 & = 3 \\ - 1 \frac{1}{4} + 1 \frac{1}{4} - \frac{b}{2} + 1 & = 3 \\ - \frac{b}{2} & = 2 \end{gathered}$

$b=-4$