Polynomials - linear combinations of powers of x

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. $(2x-3)\left(3x^2-4x+1\right)=6x^3-8x^2+2x-9x^2+12x-3=6x^3-17x^2+14x-3$

Factorising Cubics – you may be asked 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. It can be shown 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