# Thread:

1. ## Easiest way to factor quadratic equations/polynomials

I'm told that the "British" method is the most foolproof without having to continuously guess+check, and is very logical because I have trouble finding common factors and stuff in my head.

So for this problem:

x^2 + 11x + 12

I multiply 12 and 1 as per the method, and I get 12. Now I need to find two numbers that are multiples of 12 that add up to 11, and I can't find this.

So is there another different method that I can definitely get the answer with, foolproof without having to guess and check multiples as well? And what if ax^n, bx, and c are huge numbers? It'd be even more difficult using either British method or finding the GCF, right? And what if b or c is 0? Then the British method won't work at all.

2. ## Re: Easiest way to factor quadratic equations/polynomials

Yes, indeed so maybe the quadratic formula can be useful.

3. ## Re: Easiest way to factor quadratic equations/polynomials

Hi daigo!

The standard method to solve a quadratic equation is by use of the quadratic formula.

The solution of
$ax^2+bx+c=0$
is
$x={-b \pm \sqrt{b^2 - 4ac} \over 2a}$

The part under the square root $b^2 - 4ac$ is called the discriminant.
If it is negative there are no solutions.

4. ## Re: Easiest way to factor quadratic equations/polynomials

Ah, thanks. We didn't learn the quadratic equation yet, I just jumped to a bunch of questions while working on the problems that didn't require the quadratic formula I guess.

5. ## Re: Easiest way to factor quadratic equations/polynomials

In fact, most quadratics cannot be factored with integer (or rational) coefficients.

6. ## Re: Easiest way to factor quadratic equations/polynomials

Originally Posted by daigo
I'm told that the "British" method is the most foolproof without having to continuously guess+check, and is very logical because I have trouble finding common factors and stuff in my head.

So for this problem:

x^2 + 11x + 12

I multiply 12 and 1 as per the method, and I get 12. Now I need to find two numbers that are multiples of 12 that add up to 11, and I can't find this.

So is there another different method that I can definitely get the answer with, foolproof without having to guess and check multiples as well? And what if ax^n, bx, and c are huge numbers? It'd be even more difficult using either British method or finding the GCF, right? And what if b or c is 0? Then the British method won't work at all.
If this method you describe fails, then to factorise the quadratic (if it factorises, not all quadratics do) you need to complete the square, then, if possible, use the Difference of Two Squares.

$\displaystyle \begin{align*} x^2 + 11x + 12 &= x^2 + 11x + \left(\frac{11}{2}\right)^2 - \left(\frac{11}{2}\right)^2 + 12 \\ &= \left(x + \frac{11}{2}\right)^2 - \frac{121}{4} + \frac{48}{4} \\ &= \left(x + \frac{11}{2}\right)^2 - \frac{73}{4} \\ &= \left(x + \frac{11}{2}\right)^2 - \left(\frac{\sqrt{73}}{2}\right)^2 \\ &= \left(x + \frac{11}{2} - \frac{\sqrt{73}}{2}\right)\left(x + \frac{11}{2} + \frac{\sqrt{73}}{2}\right) \end{align*}$
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