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August 19, 2019, 01:03:44 am

Author Topic: I have a question on binomials  (Read 86 times)  Share 

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_.aliciaaa_

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I have a question on binomials
« on: June 10, 2019, 03:44:06 pm »
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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 »
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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.)

_.aliciaaa_

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Re: I have a question on binomials
« Reply #2 on: June 10, 2019, 07:18:49 pm »
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I understand it now, thank you