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

1 Answer
Jan 1, 2018

#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 :)