# Binomial theorem check

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

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

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.

• jvignacio

#### jvignacio

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 #### jvignacio

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.
EDIT: NEVERMIND, JUST GOT IT 