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# Factor Theorem

## 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}-240-32+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 $$y-1$$ is a factor of $$f\left(y\right)=2{y}^{4}+3{y}^{2}-5y+7$$.

### Determine how to approach the problem

For $$y-1$$ 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+3-5+7 \\ & = 7 \end{align*}

### Conclusion

Since $$f\left(1\right)\ne 0$$, $$y-1$$ 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: $$f\left(x\right)={x}^{3}+{x}^{2}-9x-9$$

### Find a factor by trial and error

Try

$$f\left(1\right)={\left(1\right)}^{3}+{\left(1\right)}^{2}-9\left(1\right)-9=1+1-9-9=-16$$

Therefore $$\left(x-1\right)$$ is not a factor.

We consider the coefficients of the given polynomial and try:

$$f\left(-1\right)={\left(-1\right)}^{3}+{\left(-1\right)}^{2}-9\left(-1\right)-9=-1+1+9-9=0$$

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 $${x}^{3}+{x}^{2}-9x-9=\left(x+1\right)\left( \ldots \right)$$

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 $${x}^{3}+{x}^{2}-9x-9=\left(x+1\right)\left({x}^{2}+?x-9\right)$$

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 $$\text{0}$$.

$$\therefore f\left(x\right)=\left(x+1\right)\left({x}^{2}-9\right)$$.

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

$$f\left(1\right)={\left(1\right)}^{3}-2{\left(1\right)}^{2}-5\left(1\right)+6=1-2-5+6=0$$

Therefore $$\left(x-1\right)$$ is a factor.

### Factorise by inspection

$${x}^{3}-2{x}^{2}-5x+6=\left(x-1\right)\left( \ldots \right)$$

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 $${x}^{3}-2{x}^{2}-5x+6=\left(x-1\right)\left({x}^{2}+?x-6\right)$$

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 $$f\left(x\right)=\left(x-1\right)\left({x}^{2}-x-6\right)$$.

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.

We can factorise the last bracket as:

\begin{align*} f(x) &= \left(x-1\right)\left({x}^{2}-x-6\right) \\ &= (x - 1)(x - 3)(x + 2) \end{align*}

# Do you need more Practice?

## Factorising cubic polynomials

Exercise 5.5

Find the remainder when $$4{x}^{3}-4{x}^{2}+x-5$$ is divided by $$x + 1$$.

\begin{align*} \text{Let } a(x)&= 4{x}^{3}-4{x}^{2}+x-5 \\ a(-1) &= 4(-1)^{3}-4(-1)^{2}+(-1)-5 \\ &= -4 -4 -1-5 \\ &= -\text{14} \end{align*}

Use the factor theorem to factorise $${x}^{3}-3{x}^{2}+4$$ completely.

\begin{align*} \text{Let } a(x)&= {x}^{3}-3{x}^{2}+4 \\ a(-1) &= (-1)^{3}-3(-1)^{2}+4 \\ &= -1 -3 +4 \\ &= 0 \\ \therefore (x+1) & \text{ is a factor} \\ a(x) &= (x + 1)({x}^{2} -4x +4 ) \\ &= (x + 1)(x - 2)(x - 2) \\ &= (x + 1)(x - 2)^{2} \end{align*}

$$f\left(x\right)=2{x}^{3}+{x}^{2}-5x+2$$

Find $$f\left(1\right)$$.

\begin{align*} f(x)&= 2{x}^{3}+{x}^{2}-5x+2 \\ f(1)&= 2(1)^{3}+(1)^{2}-5(1)+2 \\ &= 2 + 1 -5 + 2 \\ &= \text{0} \end{align*}

Factorise $$f\left(x\right)$$ completely.

\begin{align*} f(1)&= 0 \\ \therefore (x - 1) & \text{ is a factor of } f(x) \\ f(x) &= (x - 1)(2x^{2} + 3x - 2) \\ &= (x-1)(2x - 1)(x + 2) \end{align*}

Use the factor theorem to determine all the factors of the following expression:

$${x}^{3}+{x}^{2}-17x+15$$
\begin{align*} \text{Let } a(x) &= {x}^{3}+{x}^{2}-17x+15 \\ a(1) &= (1)^{3}+(1)^{2}-17(1)+15 \\ &= 1 +1 -17 +15 \\ &= 0 \\ \therefore a(x) &= (x-1)(x^{2} + 2x -15) \\ &= (x-1)(x+5)(x -3) \end{align*}

Complete: If $$f\left(x\right)$$ is a polynomial and p is a number such that $$f\left(p\right)=0$$, then $$\left(x-p\right)$$ is...

a factor of $$f(x)$$