## anonymous 4 years ago (4X)^2 - (4X)^3 A: X^-1 B:12X^-1 C:16x^2 - 4X^3 D:16x^2 - 64X^3

1. anonymous

$(4x)^2 - (4x)^3 = (4x^2)(1-4x) = 16x^2(1-4x) = 16x^2 - (16\times 4)x^3 = 16x^2 - 64x^3$

2. anonymous

sorry there is a little issue$(4x)^2−(4x)^3=(4x)^2(1−4x)$ the rest of what i wrote is right

3. anonymous

Ha Can you explain that a little bit like how you got the one and started simplifying ? im kind of confused

4. anonymous

$(4x)^2 - (4x)^3 = (4x)^2(1-4x)$ because (4x)^2 is common to both (4x)^2 and -(4x)^3, since $1\times (4x)^2 = (4x)^2,$ and $-(4x)\times (4x)^2 = (4x)^3.$ $(4x)^2(1-4x) = 16x^2(1-4x)$ because (4x)^2 = (4^2)(x^2) = 16x^2. Now just multiply out.

5. anonymous

$\left( 4x ^{2} \right)=4^{2}x ^{2}=16x ^{2}$ in the same way $\left( 4x ^{3} \right)=4^{3}x ^{3}=64x ^{3}$ substitute both of them u will get$16x ^{2}-64x ^{3}$ 16x square -64x cube

6. anonymous

That makes sense guys thanks :)