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January 21, 2020, 10:20:22 am

### AuthorTopic: I have a question on binomials  (Read 186 times) Tweet Share

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#### _.aliciaaa_

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##### I have a question on binomials
« on: June 10, 2019, 03:44:06 pm »
0
in the expansion of (ax + b)6, the coefficient of x4 is 2160 and the coefficient of x5 is −576. Find a and b.
I don't understand these types of questions so I will really appreciate an explanation
thank you

#### RuiAce

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##### Re: I have a question on binomials
« Reply #1 on: June 10, 2019, 06:13:11 pm »
+1
To help build your understanding, recall that the general form of the binomial expansion is
$(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \binom{n}{3}a^{n-3}b^3 + \dots + \binom{n}{n-2}a^2 b^{n-2} + \binom{n}{n-1}ab^{n-1}+\binom{n}{n}b^n.$
The idea is that each term is of the form $\boxed{\binom{n}{k}a^{n-k}b^k}$. This expression is therefore referred to as the 'general term' of the expansion of $(a+b)^n$.

Whereas in your example, because you're dealing with $(ax+b)^6$, your general term is of the form $\boxed{\binom{6}{k} (ax)^{6-k}b^k}$. And after using your index laws, it rearranges to $\boxed{\binom{6}{k} a^{6-k}b^k x^{6-k}}$.

Since you want to find the term with $x^4$, you first set the power on $x$ equal to 4. This gives you $6-k=4$, so $k=2$. Then, the coefficient is obtained by subbing back in for $k$, but blocking out the $x$'s. So the coefficient of $x^4$ is $\boxed{\binom62 a^4b^2}$. If this is equal to 2160, then you first have
$15 a^4 b^2 = 2160.$
Note that the value of $\binom62$ was computed on the calculator.

You should have a go at repeating the procedure for $x^5$ now and finish off the question. If you're stuck, please provide extra details.

(Note that to fully finish the question, you will need simultaneous equations here.)