Factor Theorem (ii)

Well that's all very neat and tidy, you might be thinking, but what good is this theorem?

It gives you a lovely and easy way of trying to find linear factors by substituting values into the polynomial you are given. Sometimes this does not work easily but it is often worth a try for a minute or so.

It also enables you to do slightly trickier things with polynomials.

You often get questions like this (we'll use the cubic from the previous post for this). A polynomial, P(x) = x³ + ax² + bx + 6, has (x+1) and (x+2) as factors. Find the values of a and b.

This looks tricky but it is where the factor theorem comes in.

P(-1) = 0. You know this from the theorem.

that means that (-1) + a(1) + b(-1) + 6 = 0

so -1 + a - b + 6 = 0

therefore a - b = -5

P(-2) = 0.

So -8 + 4a -2b + 6 = 0

therefore 4a - 2b = 2

You now have two equations to solve SIMULTANEOUSLY!!

a - b = -5
4a - 2b = 2

Times the first equation by 2:

2a - 2b = -10
4a - 2b = 2


SUBTRACT THE TWO TO ELIMINATE b

-2a = -12

a = 6

Plug this back into one of the equations:

2(6) - 2b = -10
-2b = -22

b = 11

a = 6, b = 11

There is nothing that hard here and yet in just writing this post I have made LOADS of mistakes, which is why it's taken me an hour or so to write.

Here are the errors I made:

1) Using the wrong constant value - ie reading the cubic wrong;
2) Forgetting to cube or square the values in the cubic equation;
3) subtracting with minus signs incorrectly;
4) multiplying incorrectly.

All because I was surfing the blogs while writing this post!