You have probably learnt how to factorise a quadratic expression: for example, *x*^{2} – 2*x* – 8 factorises to give (*x* – 4)(*x* + 2). This is useful because it enables us to solve the equation *x*^{2} – 2*x* – 8 = 0. How? Because if we use the factorised form we get (*x* – 4)(*x* + 2) = 0, and if two numbers multiply to give zero then one of the number *must* be zero. So either *x* – 4 = 0, in which case *x* = 4; or *x* + 2 = 0, in which case *x* = -2. We can verify these solutions by substituting them into the original equation:

When *x* = 4, *x*^{2} – 2*x* – 8 = 4^{2} – 2 × 4 – 8 = 16 – 8 – 8 = 0

When *x* = -2, *x*^{2} – 2*x* – 8 = (-2)^{2} – 2 × (-2) – 8 = 4 + 4 – 8 = 0

These solutions, known as the *roots* of the equation, also tell us that the graph of *y* = *x*^{2} – 2*x* – 8 intersect the *x*-axis at (-2, 0) and (4, 0).

It would be useful if we could factorise cubic expressions and above in the same way, thus enabling us to solve higher degree equations and draw the graphs. (Note: none of these expressions, including quadratics, necessarily factorise – it depends on the numbers – but if they do it saves a lot of head-scratching)! The *Factor Theorem* is a technique we can use to test whether (*ax – b*) is a factor of a polynomial. Let me show you how it works.

Consider the cubic P(*x*) = *x*^{3} + 2*x*^{2} – 11*x* – 12. Now, I’ve chosen this because I know it factorises to (*x* + 1)(*x* – 3)(*x* + 4). So:

P(*x*) = *x*^{3} + 2*x*^{2} – 11*x* – 12 = (*x* + 1)(*x* – 3)(*x* + 4)

Now let’s consider the root *x* = 3. If I substitute *x* = 3 into the right hand side I get 4 × 0 × -4 = 0. Therefore I must also get 0 when I substitute *x* = 3 into the left hand side: 27 + 18 – 33 – 12 = 0. In other words, P(3) = 0. Let’s turn this around: since P(3) = 0, it follows that *x* – 3 must be a factor. We can generalise this to: if P(*a*) = 0, then *x* – *a* is a factor of P(*x*). You might like to try this out by proving that *x* + 2 is a factor of *x*^{3} + 3*x*^{2} – 4*x* – 12 by substituting *x = *-2.

Once you have found a factor, there are various methods for dividing that factor into the polynomial to find the other factors. In this case* x*^{3} + 3*x*^{2} – 4*x* – 12 = (*x* + 2)(*x* – 2)(*x + *3)but I shall look at these methods in another blog.

So the blue bit above is the factor theorem. Nearly. We can refine it so that we can also test for factors such as (2*x* – 3). The full factor theorem states that:

If P(*b/a*) = 0, then *ax* – *b* is a factor of P(*x*)

For example, use the factor theorem to prove that 2*x* – 3 is a factor of

2*x*^{3} + 3*x*^{2} – 5*x* – 6. Do this by substituting *x* = 1.5 into the polynomial – you should get 0.

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