1. ## Binomial theorem check

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

Find the coefficient of $\displaystyle x^5$ in $\displaystyle (x^4 + 4)(3x-4)^9$

My solution

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

1. $\displaystyle x^4(3x-4)^9$
2. $\displaystyle 4(3x-4)^9$

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

So: $\displaystyle \binom{9}{9-i}$ $\displaystyle (3x)^{9-i}$ $\displaystyle (-4)^i$

now:

1. $\displaystyle \binom{9}{8}$ $\displaystyle (3)^8$ $\displaystyle (-4)^1$ = $\displaystyle 9 \times 6561 \times (-4)$

2. $\displaystyle 4 \binom{9}{4}$ $\displaystyle (3)^4$ $\displaystyle (-4)^5$ = $\displaystyle 4 \times 126 \times 81 \times (-1024)$

Solution: $\displaystyle 9 \times 6561 \times (-4) + 4 \times 126 \times 81 \times (-1024)$

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 $\displaystyle (x^4 + 4)(3x-4)^9$

My solution

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

1. $\displaystyle x^4(3x-4)^9$
2. $\displaystyle 4(3x-4)^9$

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

So: $\displaystyle \binom{9}{9-i}$ $\displaystyle (3x)^{9-i}$ $\displaystyle (-4)^i$

now:

1. $\displaystyle \binom{9}{8}$ $\displaystyle (3)^8$ $\displaystyle (-4)^1$ = $\displaystyle 9 \times 6561 \times (-4)$

2. $\displaystyle 4 \binom{9}{4}$ $\displaystyle (3)^4$ $\displaystyle (-4)^5$ = $\displaystyle 4 \times 126 \times 81 \times (-1024)$

Solution: $\displaystyle 9 \times 6561 \times (-4) + 4 \times 126 \times 81 \times (-1024)$
Your method is fine, but you have found the coefficient of $\displaystyle x^8$ in (1) and $\displaystyle x^4$ in (2).

You need to write the terms the other way around:
$\displaystyle \big((-4)+3x\big)^9$
The term in $\displaystyle x^i$ is now:
$\displaystyle \binom{9}{9-i}(-4)^{9-i}x^i$
Now you can put $\displaystyle i = 1$ and $\displaystyle 5$ respectively.

Hello jvignacioYour method is fine, but you have found the coefficient of $\displaystyle x^8$ in (1) and $\displaystyle x^4$ in (2).

You need to write the terms the other way around:
$\displaystyle \big((-4)+3x\big)^9$
The term in $\displaystyle x^i$ is now:
$\displaystyle \binom{9}{9-i}(-4)^{9-i}x^i$
Now you can put $\displaystyle i = 1$ and $\displaystyle 5$ respectively.

Ahh daym! no wonder I keep getting them wrong.. So its the other way around thank you Grandad :P

Hello jvignacioYour method is fine, but you have found the coefficient of $\displaystyle x^8$ in (1) and $\displaystyle x^4$ in (2).

You need to write the terms the other way around:
$\displaystyle \big((-4)+3x\big)^9$
The term in $\displaystyle x^i$ is now:
$\displaystyle \binom{9}{9-i}(-4)^{9-i}x^i$
Now you can put $\displaystyle i = 1$ and $\displaystyle 5$ respectively.

Hey Grandad, how do we know how what coefficient to get from 1 and 2.. In this example.. $\displaystyle x^5(x^3+5)(5x+3)^8$, when you split the two brackets, you have to find the coeff of $\displaystyle x^2$ in the first bracket then find the coeff of $\displaystyle x^5$ in the second bracket.. but in the previous example it was $\displaystyle x$ from 1 and $\displaystyle x^5$ from 2.