# 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 $\displaystyle x^5$ in $\displaystyle (3 - 4x)(2x+3)^7$

My solution

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

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

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

So now:

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

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

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

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 $\displaystyle x^5$ in $\displaystyle (3 - 4x)(2x+3)^7$

My solution

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

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

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

So now:

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

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

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