Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

\[(4x)^2 - (4x)^3 = (4x)^2(1-4x)\]
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.

Answer 5

\[\left( 4x ^{2} \right)=4^{2}x ^{2}=16x ^{2} \] 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

Answer 6

That makes sense guys thanks :)