# Thread: Binomial theorem question check #2

1. ## Binomial theorem question check #2

hey guys, just need to check if my working out and solution is correct.

Find the coefficient of $x^5$ in $(3 - 4x)(2x+3)^7$

My solution

So I started off by separating both brackets into 2 separate equations.

1. $3(2x+3)^7$
2. $-4x(2x+3)^7$

Now I find the coefficient of $x^5$ in equation 1 and I find the coefficient of $x^4$ in equation 2.

So now:

1. $3 \binom{7}{5}$ $(3)^2$ $(2)^5$ = $3 \times 21 \times 9 \times 32$

2. $-4 \binom{7}{4}$ $(3)^3$ $(2)^4$ = $-4 \times 35 \times 27 \times 16$

Solution: $3 \times 21 \times 9 \times 32 - 4 \times 35 \times 27 \times 16$

2. Hello jvignacio
Originally Posted by jvignacio
hey guys, just need to check if my working out and solution is correct.

Find the coefficient of $x^5$ in $(3 - 4x)(2x+3)^7$

My solution

So I started off by separating both brackets into 2 separate equations.

1. $3(2x+3)^7$
2. $-4x(2x+3)^7$

Now I find the coefficient of $x^5$ in equation 1 and I find the coefficient of $x^4$ in equation 2.

So now:

1. $3 \binom{7}{5}$ $(3)^2$ $(2)^5$ = $3 \times 21 \times 9 \times 32$

2. $-4 \binom{7}{4}$ $(3)^3$ $(2)^4$ = $-4 \times 35 \times 27 \times 16$

Solution: $3 \times 21 \times 9 \times 32 - 4 \times 35 \times 27 \times 16$