Cubic Polynomial

Solution in detail

$x^{3}-12x^{2}+48x-64=0$

We know that $-12x^{2}=-4x^{2}-8x^{2}$ and $48x=32x+16x$ , substitute inside the equation we will get

$x^{3}-4x^{2}-8x^{2}+32x+16x-64=0$

divide this equation into 3 parts which is

$(x^{3}-4x^{2}) + (-8x^{2}+32x) + (16x-64) = 0$

Factor 1 part by 1 part

We will get $x^2(x-4)-8x(x-4)+16(x-4)=0$

Factor out $x-4$

$(x^2-8x+16)(x-4)=0$

Factor the quadratic equation in front

$x^2-2(4)x+4^2$ , we know that $(a-b)^2=a^2+b^2-2ab$

So, $(x-4)(x-4)(x-4)=0$ which is $(x-4)^{3}=0$

The only root is $x=4$

Share your solution if you solved this too :)

Apply the substitution $x=y-\frac{b}{3a}=y-\frac{-12}{3(1)}=y+4$ and simplify to get $y^3=0$ which has $0$ as a root with multiplicity 3.So one root of the equation is $x=y+4=0+4=4$ meaning $x-4$ is a factor.Upon dividing we get $x^2-8x+16$ which can be factored as $(x-4)^2$ so the cubic can be factored as $(x-4)^3=0$ therefore the equation has one real root with multiplicity 3 which is $\boxed{4}$