# How do you use the binomial theorem to expand and simplify the expression 3(x+1)^5-4(x+1)^3?

Mar 23, 2017

The result is ${x}^{5} + 5 {x}^{4} + 26 {x}^{3} + 18 {x}^{2} + 3 x - 1$

Details of the expansions follow...

#### Explanation:

The first binomial will expand into

$a {x}^{5} + b {x}^{4} + c {x}^{3} + {\mathrm{dx}}^{2} + f x + g$

The trick is coming up with $a , b , c , d , f \mathmr{and} g$ in as painless a way as possible. You can use Pascal's triangle, or combinations ""_nC_r, where $n$ is 5, and $r$ is 0, 1, 2, 3, 4, 5 in that order.

Either way, the coefficients turn out to be 1, 5, 10, 10, 5, 1 in that order.

${x}^{5} + 5 {x}^{4} + 10 {x}^{3} + 10 {\mathrm{dx}}^{2} + 5 x + 1$

Now multiply the whole thing by 3.

$3 {x}^{5} + 15 {x}^{4} + 30 {x}^{3} + 30 {x}^{2} + 15 x + 3$

Subtract from this the second expanded binomial (also get the coefficients in one of the ways mentioned above):

$4 \left({x}^{3} + 3 {x}^{2} + 3 x + 1\right)$

$4 {x}^{3} + 12 {x}^{2} + 12 x + 4$

The finished result is found by combining like terms (those in x^3, x^2, x, etc.)

${x}^{5} + 5 {x}^{4} + 26 {x}^{3} + 18 {x}^{2} + 3 x - 1$

Hope that all makes sense!