Explain this: I read that to factorize you need to find numbers that adds or minus to the middle value and multiply to give the last value. But how did this come up? The middle quantities when multiplied don't make up 12. Please explain this.
#12^2-7y-12# =0
Then:
#12^2-16y+9y-12# =0
Then:
1 Answer
What you start describing is appropriate for monic quadratics, but if the leading coefficient is not
Explanation:
If you have a monic quadratic (i.e. leading coefficient is
#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alpha beta#
If the leading coefficient is not
One way of simplifying things a bit is an "AC" method.
In the example:
#12x^2-7x-12 = 0#
we multiply the first coefficient
Then noting that the sign of the constant term is
(If the sign on the constant term was
The pair
In order to find this pair you can reason as follows:
The prime factorisation of
#144 = 2^4 * 3^2#
Note that
So if we split
#144 * 1" "# giving difference#144-1 = 143#
#16 * 9" "# giving difference#16-9 = 7# as required.
We then use this pair to split the middle term and factor by grouping:
#0 = 12x^2-7x-12#
#color(white)(0) = (12x^2-16x)+(9x-12)#
#color(white)(0) = 4x(3x-4)+3(3x-4)#
#color(white)(0) = (4x+3)(3x-4)#
Hence roots
Another way we might have approached this quadratic would be to divide it by
Footnotes
To understand why the AC method works, consider a quadratic equation:
#ax^2+bx+c = 0#
Multiplying by
#0 = a(ax^2+bx+c)#
#color(white)(0) = a^2x^2+abx+ac#
#color(white)(0) = (ax)^2+b(ax)+ac#
#color(white)(0) = x_1^2+bx_1+ac#
where
Having arrived at a monic quadratic in
This will give us a factorisation of