Originally Posted by

**Soroban** Hello, badandy328!

Do you **know** the Binomial Theorem?

. . If you do, exactly *where* is your difficulty?

$\displaystyle (4x-2)^5\;=\;\binom{5}{5}(4x)^5(-2)^0 + \binom{5}{4}(4x)^5(-2)^1 + \binom{5}{3}(4x)^3(-2)^2$

. . . . . . . . $\displaystyle + \binom{5}{2}(4x)^2(-2)^3 + \binom{5}{1}(4x)^1(-2x)^4 + \binom{5}{0}(4x)^0(-2)^5$

. . $\displaystyle = \;1(1024x^5)(1) + 5(256x^4)(-2) + 10(64x^3)(4) +$ $\displaystyle 10(16x^2)(-8) + 5(4x)(16) + 1(1)(-32) $

. . $\displaystyle = \;1024x^5 - 2560x^4 + 2560x^3 - 1280x^2 + 210x - 32$