# What is the standard form of y=8(x - 1) (x^2 +6)(x^3 + 8) ?

Jan 1, 2018

$y = 8 {x}^{6} - 8 {x}^{5} + 48 {x}^{4} + 16 {x}^{3} - 64 {x}^{2} + 384 x - 384$

#### Explanation:

There are many ways to expand this polynomial. The way I did it is as follows:

Step One
Expand the last two brackets;
$\left({x}^{2} + 6\right) \left({x}^{3} + 8\right) = {x}^{5} + 6 {x}^{3} + 8 {x}^{2} + 48$

Step Two
Multiply everything by 8;
$8 \left({x}^{2} + 6\right) \left({x}^{3} + 8\right) = 8 \left({x}^{5} + 6 {x}^{3} + 8 {x}^{2} + 48\right)$
$8 \left({x}^{5} + 6 {x}^{3} + 8 {x}^{2} + 48\right) = 8 {x}^{5} + 48 {x}^{3} + 64 {x}^{2} + 384$

Step Three
Multiply by $\left(x - 1\right)$
$8 \left(x - 1\right) \left({x}^{2} + 6\right) \left({x}^{3} + 8\right) = 8 \left(x - 1\right) \left({x}^{5} + 6 {x}^{3} + 8 {x}^{2} + 48\right)$
$8 \left(x - 1\right) \left({x}^{5} + 6 {x}^{3} + 8 {x}^{2} + 48\right) = \left(x - 1\right) \left(8 {x}^{5} + 48 {x}^{3} + 64 {x}^{2} + 384\right)$
$\left(x - 1\right) \left(8 {x}^{5} + 48 {x}^{3} + 64 {x}^{2} + 384\right) = 8 {x}^{6} - 8 {x}^{5} + 48 {x}^{4} + 16 {x}^{3} - 64 {x}^{2} + 384 x - 384$

Hope that helps :)