Question

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

Answer 1

`y=8x^6-8x^5+48x^4+16x^3-64x^2+384x-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;
`(x^2+6)(x^3+8)=x^5+6x^3+8x^2+48`

Step Two
Multiply everything by 8;
`8(x^2+6)(x^3+8)=8(x^5+6x^3+8x^2+48)`
`8(x^5+6x^3+8x^2+48)=8x^5+48x^3+64x^2+384`

Step Three
Multiply by `(x-1)`
`8(x-1)(x^2+6)(x^3+8)=8(x-1)(x^5+6x^3+8x^2+48)`
`8(x-1)(x^5+6x^3+8x^2+48)=(x-1)(8x^5+48x^3+64x^2+384)`
`(x-1)(8x^5+48x^3+64x^2+384)=8x^6-8x^5+48x^4+16x^3-64x^2+384x-384`

Hope that helps :)