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Factorising polynomials
By Nicola Campbell (P3008) on Wednesday, October 4, 2000 - 07:54 pm:

Hello! I'm Nicola, and I'm having a problem with factorising polynomials. I understand how to do basic factorising, and breaking down quadratic equations, but I don't understand:
3x3-27x2+81x-81

Thankyou!

By Anonymous on Wednesday, October 4, 2000 - 08:18 pm:

3(x-3)3

By Dan Goodman (Dfmg2) on Wednesday, October 4, 2000 - 10:02 pm:

A couple of tips to bear in mind when factorising polynomials. First, take out any common factors you can, for instance, you can write your polynomial as 3(x3-9x2+27x-27). This also works for a polynomial like x3+5x2, which can be written as x2(x+5).

Once you've taken out any common factors, you can try and factorise whatever is left, which is often easier. Sometimes you can just spot what it is (the fancy word for this is factorising "by inspection"), but often you can't. Another trick is to look for values which make the polynomial equal to zero.

For instance, in the polynomial you've given, put x=3 to get 3×33-27×32+81×3-81=0. If x=a makes the polynomial equal to zero, then (x-a) is a factor of the polynomial. You can find values which make the polynomial equal to zero by trial and error usually, just try x=-1,1,2,-2,3,-3,4,-4, etc. For your polynomial, x=3 is a solution, so (x-3) is a factor.

Once you know this, you can simplify the polynomial even further. You know that you can write your polynomial as 3(x-3)(some polynomial). Also, the "some polynomial" bit will be a quadratic, because the x term in the (x-3) factor will be multiplied by the x2 in the "some polynomial" bit to give the x3 in your polynomial. So, you can write your polynomial as 3(x-3)(ax2+bx+c) for some numbers a, b and c (these numbers are called coefficients). The original polynomial has a 3x3 term in it, and the only x3 term in the new polynomial is 3ax3, so a must be equal to 1, so that 3x3=3ax3. You can continue in this way to work out what b and c are as well, it turns out that your polynomial comes out to be 3(x-3)(x2-6x+9). Now you can continue in this way, because you can then factorise x2-6x+9 using the same method, and you get 3(x-3)(x-3)(x-3)=3(x-3)3. Hope that helps.

By Thomas Mooney (P3048) on Tuesday, October 10, 2000 - 02:30 pm:

An easier way to evaluate the zeros is to write the polynomial in nested form, this makes the arithmetic easier. For 3x3-27x2+81x-81, write this as x(3x2-27x+81)-81=x(x(3x-27)+81)-81 and then sub your values in to it.