# Binomial theorem question check #2

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

MHF Hall of Honor
Hello 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: $$\displaystyle 3 \times 21 \times 9 \times 32 - 4 \times 35 \times 27 \times 16$$