Hi
I can't see how to factorise this. And can't think of another method. Can someone please help?
x^3 - 12x + 16 = 0
Angus
This polynomial is $\displaystyle \displaystyle P(x) = x^3 - 12x + 16$.
Notice that $\displaystyle \displaystyle P(2) = 2^3 - 12\cdot 2 + 16 = 8 - 24 + 16 = 0$. So $\displaystyle \displaystyle (x - 2)$ is a factor.
Long divide to find the quadratic factor and factorise the quadratic if possible.
Another way....
$\displaystyle 12x-x^3=16\Rightarrow\ x\left(12-x^2\right)=16$
16 factorises to 16(1)=8(2)=4(4) so we can quickly discover
whether or not the function has an integer root.
If it does, the factoring is made simple.
$\displaystyle x=2\Rightarrow\ 2(12-4)=2(8)$ works
$\displaystyle x=4$ gives $\displaystyle -16,$ but $\displaystyle x=-4$ gives $\displaystyle (-4)(-4)=16$
$\displaystyle x^3-12x+16=(x-2)(x+4)(x+c)\Rightarrow\ -8c=16\Rightarrow\ c=-2$
Check that this holds.
Or
$\displaystyle (x-2)\left(x^2+bx+c\right)=x^3-12x+16\Rightarrow\ c=-8,\;\;bx^2-2x^2=0\Rightarrow\ b=2$
as there is no $\displaystyle x^2$ term.
Archie Meade's method assumes that the roots will be integer since those are the only factors he is trying.
Another method is to use the "rational root theorem"- that if m/n is a rational number root of a polynomial with integer coefficients then the denominator, n, must divide the leading coefficient and the numerator, m, must divide the constant term. Here, the leading coefficent is 1 so the denominator must be 1 or -1 (any rational root must be an integer just as Archie Meade thought) and the constant term is 16 so any rational root must evenly divide 16: 1, -1, 2, -2, 4, -4, 8, -8, 16,or -16 are possible roots. Trying each of those numbers in the polynomial we see that $\displaystyle (2)^3- 12(2)+ 16= 8- 24+ 16= 0$ and that $\displaystyle (-4)^3- 12(-4)+ 16= -64+ 48+ 16= 0$
Using x-2 and x+ 4 as factors, we see that the third factor is x- 2 so that x= 2 is a double root.
Of course, the rational root theorem can only point out possible rational roots. It may be that a polynomial has no rational number roots but in that case, no method is going to be easy!