Binomial theorem question check #2

Oct 2007
517
6
Santiago
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. \(\displaystyle -4 \binom{7}{4}\) \(\displaystyle (3)^3\) \(\displaystyle (2)^4\) = \(\displaystyle -4 \times 35 \times 27 \times 16\)

So now adding both up:

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

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
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. \(\displaystyle -4 \binom{7}{4}\) \(\displaystyle (3)^3\) \(\displaystyle (2)^4\) = \(\displaystyle -4 \times 35 \times 27 \times 16\)

So now adding both up:

Solution: \(\displaystyle 3 \times 21 \times 9 \times 32 - 4 \times 35 \times 27 \times 16\)
Looks good to me!

Grandad