Hate Algebraic Long Division?

So here goes my first post on this blog, and I thought I’d talk about one of the biggest causes of confusion in students I have tutored in the past. This is the complicated and unintuitive method of factorising polynomials by using algebraic long division. Most of the time it is students who are preparing for the Edexcel core modules who use this method, since the main texbook for this exam board uses the method.

Before I get started on an example showing the alternative let me just add a bit of a disclaimer: I do not have any issue with pupils using algebraic long division if it works for them, use it if you are fine with it! The alternative method of equating coefficients is not really any quicker. I personally prefer it as to me it seems to solve the problem logically and Mathematically, as opposed to the long division method which gives no understanding whatsoever of the underlying principles but forces you to simply memorise a complicated method (seems old fashioned to me!)

Anyway,  so if you do not want to touch algebraic long division EVER AGAIN, here is an explanation of the alternative method of equating coefficients (obviously if you already know this method there is not much need for you to read on – unless you’d like to offer a clearer way for  me to describe it!)

Lets lead with an example (OCR Jan 07 C2):

8. The polynomial f(x) is defined by f(x) = x³ − 9x² + 7x + 33.


(i) Find the remainder when f(x) is divided by (x + 2). [2]
(ii) Show that (x − 3) is a factor of f(x). [1]
(iii) Solve the equation f(x) = 0, giving each root in an exact form as simply as possible. [6]

(i) , (ii) I’ll leave these parts to the reader, since they are not the focus of this post. (Though if you need a hint, use the remainder theorem for part (i) and the factor theorem for part (ii) – which, incidentally are the SAME theorem – the factor theorem is simply a special case of the remainder theorem; if the remainder is 0, you have a factor!)

(iii) Solving cubic equations is difficult! There is no nice easy-to-remember formula like there is with quadratics. (There IS a formula for cubics – it was discovered by some Italian Mathematicians in the Renaissance, here’s a link to the full formula http://www.math.vanderbilt.edu/~schectex/courses/cubic/ )

So the first thing to do is find a factor by inspired guesswork, plain guesswork, divination or alchemy – then you need to find the other factor by either dividing the polynomial by the factor you have (errk!) or use the following method:

Since (x-3) is a factor of f(x), we can write f(x) as;

f(x) = x³ − 9x² + 7x + 33 = (x-3)(Something)

Now, this ‘Something’ is the other factor we need to find. We do not know what the ‘something’ is, but since f(x) is a cubic it must be a quadratic (take a moment to convince yourself of this!) The general form of a quadratic is Ax²+Bx+C

So,

‘Something’ = Ax²+Bx+C

I.e.  f(x) = x³ − 9x² + 7x + 33 = (x-3)(Ax²+Bx+C)
All we need to do now is find A, B and C and we have our second factor! This can be done very quickly and just by inspection once you become proficient – (really, you can work out A B and C in your head, start with A and C they are easy, then work out B!) –  however, lets take this one step at a time.

First expand the right hand side:

(x-3)(Ax²+Bx+C) = Ax³+Bx²+Cx – 3Ax² – 3Bx – 3C

= Ax³ + (B-3A)x² + (C – 3B)x – 3C
(The second line is found by collecting the x and x squared terms together)

Here’s where the Equating Coefficients method is used. So, we now have;

x³ − 9x² + 7x + 33 = Ax³ + (B-3A)x² + (C – 3B)x – 3C

On both sides of the equation, the coefficients of x³, x², x and the constant must be the same (again, take a moment to convince yourself of this!) This is the fact that will tell us what A, B and C are;

1− 9+ 7x + 33 = Ax³ + (B-3A)x² + (C – 3B)x – 3C
This gives us four new equations (each colour in the above equation is equal,) so let’s solve these one at a time;

First 1 = A (No solving required)

Second -9 = B – 3A (Since A = 1, this gives us B – 3 = -9)

B = -6
Third 7 = C – 3B (We know B = -6, so C + 18 = 7)

C = -11
Fourth 33 = -3C (We don’t really NEED to solve this, but it gives us a useful check that we solved the earlier equations correctly)

DONE! So we now have the second factor by putting our values for A, B and C back into our equation:

f(x) = x³ − 9x² + 7x + 33 = (x-3)(x²-6x-11)
The completion of this question is now left to the reader. (One solution of f(x) = 0 is clearly 3, the other two are found by solving x²-6x-11 = 0)

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