Find the remainder when \(4{x}^{3}4{x}^{2}+x5\) is divided by \(x + 1\).
Factor Theorem
Previous
Remainder theorem

Next
Solving cubic equations

5.4 Factor theorem (EMCGW)
If an integer \(a\) is divided by an integer \(b\), and the answer is \(q\) with the remainder \(r = 0\), then we know that \(b\) is a factor of \(a\).
\begin{align*} a &= b \times q + r \\ \text{and } r &= 0 \\ \text{then we know that } a &= b \times q \\ \text{and also that } \frac{a}{b} &= q \end{align*}This is also true of polynomials; if a polynomial \(a(x)\) is divided by a polynomial \(b(x)\), and the answer is \(Q(x)\) with the remainder \(R(x) = 0\), then we know that \(b(x)\) is a factor of \(a(x)\).
\begin{align*} a(x) &= b(x) \cdot Q(x) + R(x) \\ \text{and } R(x) &= 0 \\ \text{then we know that } a(x) &= b(x) \cdot Q(x) \\ \text{and also that } \frac{a(x)}{b(x)} &= Q(x) \end{align*}The factor theorem describes the relationship between the root of a polynomial and a factor of the polynomial.
The Factor theorem
If the polynomial \(p(x)\) is divided by \(cx  d\) and the remainder, given by \(p \left( \frac{d}{c} \right),\) is equal to zero, then \(cx  d\) is a factor of \(p(x)\).
Converse: if \((cx  d)\) is a factor of \(p(x)\), then \(p \left( \frac{d}{c} \right) = 0\).
Worked example 9: Factor theorem
Using the factor theorem, show that \(y+4\) is a factor of \(g\left(y\right)=5{y}^{4}+16{y}^{3}15{y}^{2}+8y+16\).
Determine how to approach the problem
For \(y+4\) to be a factor, \(g\left(4\right)\) must be equal to \(\text{0}\).
Calculate \(g\left(4\right)\)
\begin{align*} g\left(y\right) & = 5{y}^{4}+16{y}^{3}15{y}^{2}+8y+16 \\ \therefore g\left(4\right) & = 5{\left(4\right)}^{4}+16{\left(4\right)}^{3}15{\left(4\right)}^{2}+8\left(4\right)+16 \\ & = 5\left(256\right)+16\left(64\right)15\left(16\right)+8\left(4\right)+16 \\ & = \text{1 280}\text{1 024}24032+16 \\ & = 0 \end{align*}Conclusion
Since \(g\left(4\right)=\text{0}\), \(y+4\) is a factor of \(g\left(y\right)\).
In general, to factorise a cubic polynomial we need to do the following:
 Find one factor by trial and error: consider the coefficients of the given cubic polynomial \(p(x)\) and guess a possible root (\(\frac{c}{d}\)).
 Use the factor theorem to confirm that \(\frac{c}{d}\) is a root; show that \(p\left(\frac{c}{d}\right)\) = 0.
 Divide \(p(x)\) by the factor (\(cx  d\)) to obtain a quadratic polynomial (remember to be careful with the signs).
 Apply the standard methods of factorisation to determine the two factors of the quadratic polynomial.
Worked example 10: Factor theorem
Use the factor theorem to determine if \(y1\) is a factor of \(f\left(y\right)=2{y}^{4}+3{y}^{2}5y+7\).
Determine how to approach the problem
For \(y1\) to be a factor, \(f\left(1\right)\) must be equal to \(\text{0}\).
Calculate \(f\left(1\right)\)
\begin{align*} f\left(y\right) & = 2{y}^{4}+3{y}^{2}5y+7 \\ \therefore f\left(1\right) & = 2{\left(1\right)}^{4}+3{\left(1\right)}^{2}5\left(1\right)+7 \\ & = 2+35+7 \\ & = 7 \end{align*}Conclusion
Since \(f\left(1\right)\ne 0\), \(y1\) is not a factor of \(f\left(y\right)=2{y}^{4}+3{y}^{2}5y+7\).
Worked example 11: Factorising cubic polynomials
Factorise completely: \(f\left(x\right)={x}^{3}+{x}^{2}9x9\)
Find a factor by trial and error
Try
\(f\left(1\right)={\left(1\right)}^{3}+{\left(1\right)}^{2}9\left(1\right)9=1+199=16\)Therefore \(\left(x1\right)\) is not a factor.
We consider the coefficients of the given polynomial and try:
\(f\left(1\right)={\left(1\right)}^{3}+{\left(1\right)}^{2}9\left(1\right)9=1+1+99=0\)
Therefore \(\left(x+1\right)\) is a factor, because \(f\left(1\right)=0\).
Factorise by inspection
Now divide \(f\left(x\right)\) by \(\left(x+1\right)\) using inspection:
Write \({x}^{3}+{x}^{2}9x9=\left(x+1\right)\left( \ldots \right)\)
The first term in the second bracket must be \({x}^{2}\) to give \({x}^{3}\) and make the polynomial a cubic.
The last term in the second bracket must be \(\text{9}\) because \((+1)(9)=9\).
So we have \({x}^{3}+{x}^{2}9x9=\left(x+1\right)\left({x}^{2}+?x9\right)\)
Now, we must find the coefficient of the middle term:
\(\left(+1\right)\left({x}^{2}\right)\) gives the \({x}^{2}\) in the original polynomial. So, the coefficient of the \(x\)term must be \(\text{0}\).
\(\therefore f\left(x\right)=\left(x+1\right)\left({x}^{2}9\right)\).
Write the final answer
We can factorise the last bracket as a difference of two squares:
\begin{align*} f(x) &= \left(x+1\right)\left({x}^{2}9\right) \\ &= (x + 1)(x  3)(x + 3) \end{align*}Worked example 12: Factorising cubic polynomials
Use the factor theorem to factorise \(f(x) = {x}^{3}2{x}^{2}5x+6\).
Find a factor by trial and error
Try
\(f\left(1\right)={\left(1\right)}^{3}2{\left(1\right)}^{2}5\left(1\right)+6=125+6=0\)Therefore \(\left(x1\right)\) is a factor.
Factorise by inspection
\({x}^{3}2{x}^{2}5x+6=\left(x1\right)\left( \ldots \right)\)
The first term in the second bracket must be \({x}^{2}\) to give \({x}^{3}\) if we work backwards.
The last term in the second bracket must be \(\text{6}\) because \((1)(6)=+6\).
So we have \({x}^{3}2{x}^{2}5x+6=\left(x1\right)\left({x}^{2}+?x6\right)\)
Now, we must find the coefficient of the middle term:
\(\left(1\right)\left({x}^{2}\right)\) gives \({x}^{2}\). So, the coefficient of the \(x\)term in the second bracket must be \(\text{1}\) to give another \({x}^{2}\) so that overall we have \({x}^{2} {x}^{2} = 2{x}^{2}\).
So \(f\left(x\right)=\left(x1\right)\left({x}^{2}x6\right)\).
Make sure that the expression has been factorised correctly by checking that the coefficient of the \(x\)term also works out: \((x)(6) + (1)(x) = 5x\), which is correct.
Write the final answer
We can factorise the last bracket as:
\begin{align*} f(x) &= \left(x1\right)\left({x}^{2}x6\right) \\ &= (x  1)(x  3)(x + 2) \end{align*}Factorising cubic polynomials
Use the factor theorem to factorise \({x}^{3}3{x}^{2}+4\) completely.
\(f\left(x\right)=2{x}^{3}+{x}^{2}5x+2\)
Find \(f\left(1\right)\).
Factorise \(f\left(x\right)\) completely.
Use the factor theorem to determine all the factors of the following expression:
\({x}^{3}+{x}^{2}17x+15\)Complete: If \(f\left(x\right)\) is a polynomial and p is a number such that \(f\left(p\right)=0\), then \(\left(xp\right)\) is...
Previous
Remainder theorem

Table of Contents 
Next
Solving cubic equations
