Binomial theorem check

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

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

So now adding both up:

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

Grandad

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

So now adding both up:

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.

Grandad
 
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Reactions: jvignacio
Oct 2007
517
6
Santiago
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.

Grandad
Ahh daym! no wonder I keep getting them wrong.. So its the other way around :) thank you Grandad :p
 
Oct 2007
517
6
Santiago
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.

Grandad
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.

thanks

EDIT: NEVERMIND, JUST GOT IT :)
 
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