# limx-2 x6 -24x-16/x3+2x-12. How i can solve it by factorisation?

limx-2 x6 -24x-16/x3+2x-12
Substitute x=2 in x6-24x-16 ; Then we get
26-24(2)-16=0
So we got 0. So, x=2 is one of its root. Let us divide it by synthetic division.

The quotient here is
x5+2x4+4x3+8x2+16x+8
Similarly substitute x=2 in x3+2x-12.
23+2(2)-12=0.
So we got 0. So, x=2 is one of its root. Let us divide it by synthetic division.

The quotient here is
x2+2x+6.
Let us apply the limit now.
$\underset{x\to 2}{\mathrm{lim}}\frac{{x}^{6}-24x-16}{{x}^{3}-2x+12}\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\frac{\left(x-2\right)\left({x}^{5}+2{x}^{4}+4{x}^{3}+8{x}^{2}+16x+8\right)}{\left(x-2\right)\left({x}^{2}+2x+6\right)}\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\frac{\left({x}^{5}+2{x}^{4}+4{x}^{3}+8{x}^{2}+16x+8\right)}{\left({x}^{2}+2x+6\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left({2}^{5}+2\left({2}^{4}\right)+4\left({2}^{3}\right)+8\left({2}^{2}\right)+16\left(2\right)+8\right)}{\left(\left(2{\right)}^{2}+2\left(2\right)+6\right)}\phantom{\rule{0ex}{0ex}}=\frac{168}{14}=12$

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