# Math Help - Binomial theorem help

1. ## Binomial theorem help

hey guys, question:
Find the coefficient of $x^{-4}$ in $(4x^7-2x^{-3})^8$

I know we are ment to rewrite $(4x^7-2x^{-3})^8$ so there are two expressions to the power of 8 but im not sure how we are ment to rewrite it. I got the answer $(x^{-3})^8 (4x^{10} -2)^8$ but not sure how they got this... then it says we see that finding the coefficient of $x^{-4}$ is equivalent to finding the coefficient of $x^{20}$.. how did they get $x^{20}$ ?

thank you..

2. Originally Posted by jvignacio
hey guys, question:
Find the coefficient of $x^{-4}$ in $(4x^7-2x^{-3})^8$

I know we are ment to rewrite $(4x^7-2x^{-3})^8$ so there are two expressions to the power of 8 but im not sure how we are ment to rewrite it. I got the answer $(x^{-3})^8 (4x^{10} -2)^8$ but not sure how they got this... then it says we see that finding the coefficient of $x^{-4}$ is equivalent to finding the coefficient of $x^{20}$.. how did they get $x^{20}$ ?

thank you..
The general term is ${8 \choose r} (4x^7)^{8 - r} \left(-\frac{2}{x^3} \right)^r = {8 \choose r} 4^{8-r} (-2)^r x^{56-7r - 3r}$.

Find the value of r such that 56 - 7r - 3r = -4 and use it ot get the coefficient.

3. Originally Posted by mr fantastic
The general term is ${8 \choose r} (4x^7)^{8 - r} \left(-\frac{2}{x^3} \right)^r = {8 \choose r} 4^{8-r} (-2)^r x^{56-7r - 3r}$.

Find the value of r such that 56 - 7r - 3r = -4 and use it ot get the coefficient.
wow that totally makes more sense...! thanks mate