Question

(2x^2-4xy+2y^2)(3x-4)

A.) 6x^3-12x^2y+6xy^2-8x^2+16xy-8y^2
B.) 6x^3-6x^2y-8x^2+16xy-8y^2
C.) 2x^2-4xy+2y^2+3x-4
D.) 6x^3-12x^2y+6xy^2-8x^2-16xy+8y^2

Answer 1

Just distribute the terms!


2x^2 multiplied with 3x and -4
-4xy multiplied with 3x and -4
2y^2 multiplied with 3x and -4

(2x^2-4xy+2y^2)(3x-4)
= (2x^2 * 3x) + (2x^2 * -4) + (-4xy * 3x) + (-4xy * -4) + (2y^2 * 3x) + (2y^2 * -4)

and simplify

Answer 2

Expand by distributing sum groups.
2x^2(3x-4)-4xy(3x-4)+2y^2(3x-4)
Expand by distributing terms.
6x^3-8x^2-4xy(3x-4)+2y^2(3x-4)
Expand by distributing terms.
6x^3-8x^2-(12x^2y-16xy)+2y^2(3x-4)
Expand by distributing terms.
6x^3-8x^2-(12x^2y-16xy)+6y^2x-8y^2
Remove parentheses.
= 6x^3-8x^2-12x^2y+16xy+6y^2x-8y^2

Answer 3

That is correct!

If you re-arrange the terms a little bit, you'll get

6x^3-12x^2y+6xy^2-8x^2+16xy-8y^2

Answer 4

Exactly!

Answer 5

Ok thanks for the help!!!

Answer 6

It was my pleasure!