- The easiest way to do this is to use factor theorem. Factor theorem states that $f(x)$ is divisible by $x-2$ if $f(2)=0$. $f(2)=2^5-3(2)^3-8=32-24-8=0$

Hence $x-2$ is a factor of $f(x)$ and $f(x)$ is divisible by $x-2$. - According to Factor Theorem, if $f\left(\frac{4}{5}\right)=0$ then $5x-4$ is a factor and we can write $f(x)=(5x-4)(2x^2-18)$. We identify the quadratic factor by inspection – in order to get $10x^3$ as the first term when expanding, the first term in the quadratic must be $2x^2$. Since the expansion gives the correct $x^2$ term we don’t require an $x$ term in the quadratic. It remains to find the constant term which can be seen from the final term in $f(x)$. We factorise the quadratic term itself as $2(x-3)(x+3)$ since once a 2 is factored out, a difference of squares remains. We have $f(x)=2(5x-4)(x-3)(x+3)$.

# Polynomials

Polynomials are linear combinations of powers of x. The highest power is the order of the polynomial. For example, a cubic is a polynomial of order 3. (What does the word ‘polynomial’ mean?).

You may be required to manipulate polynomials including expanding brackets, simplifying expressions and factorising. There may also be questions involving Polynomial Division and/or with the use of Factor Theorem.

### Manipulating Polynomials

**Expanding and Simplifying**– To expand and simpify $(2x-3)\left(3x^2-4x+1\right)$ – this works in the same way as expanding double brackets but you should end up with 6 terms before simplification: $\begin{array}{l}(2x-3)\left(3x^2-4x+1\right)&=&6x^3-8x^2+2x-9x^2+12x-3\\&=&6x^3-17x^2+14x-3\end{array}$**Factorising Cubics**– they may ask you to factorise $x^3+3x^2-4$, for example. The trick is to inspect the cubic and see if you can guess a root. In this case, $x=1$ is a root since it gives $f(1)=0$. It follows that $x-1$ is a factor. We show that $x^3+3x^2-4=(x-1)(x^2+4x+4)$ by guessing the quadratic inside the brackets and expanding then improving. See the Cubics page, Example 2.2 for another example. Alternatively, you can use polynomial division as below and seen in Example 3. Fully factorised $x^3+3x^2-4=(x-1)(x^2+4x+4)=(x-1)(x+2)^2$.

### Factor Theorem & Polynomial Division

**Factor Theorem**– Factor theorem states that if a polynomial is divisible by $ax+b$ then $f\left(-\frac{b}{a}\right)=0$ and vice versa. It follows from this that if a polynomial is divisible by $ax-b$ then $f\left(\frac{b}{a}\right)=0$ and vice versa. For example, if $2x-3$ is a factor of $f(x)$, then $f\left(\frac{3}{2}\right)=0$. Similarly, if $f\left(-\frac{4}{5}\right)=0$ then $5x+4$ is a factor. In simpler cases, if $f(2)=0$, for example, then $x-2$ is a factor of $f(x)$. Simplify a problem by identifying a factor of a polynomial before you use polynomial division (see below).**Polynomial Division**– Polynomial Division combines algebra with the technique of long division. The idea is to identify the factor required for the left most term in each step. This factor then goes on top. For example, in Example 3, a cubic polynomial with a $2x^3$ term is being divided by a linear function with an $x$ term. The missing factor is thus $2x^2$. This is then multiplied by $x+3$ to see what remains to find.

## Examples of Polynomials

- Expand and simplify $(3x-1)(2x+3)(x+1)$.
- Factorise fully $ 6x^2+11x^2-x-6$.

- Start by multiplying out one pair of brackets: $(6x^2-2x+9x-3)(x+1)=(6x^2+7x-3)(x+1)$. Then multiply out the rest: $6x^3+7x^2-3x+6x^2+7x-3=6x^3+13x^2+4x-3$.
- By inspection, we can see that $x+1$ is a factor of $ f(x)=6x^3+11x^2-x-6$. This can be seen by evaluating $f(-1)$ and using Factor Theorem above. It then remains to see what the quadratic factor will be. Again, by inspection (guessing and multiplying out rather than using polynomial division) we have $6x^3+11x^2-x-6=(x+1)(6x^2+5x-6)$ Finally, we factorise the quadratic to give $6x^3+11x^2-x-6=(x+1)(2x+3)(3x-2)$.

- Show that $f(x)=x^5-3x^3-8$ is divisible by $x-2$.
- Given that $f\left(\frac{4}{5}\right)=0$, factorise fully $f(x)=10x^3-8x^2+90x+72$.