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You are given the following expression:

$$(ax + by)^{n} = -15120x^{4}y^{3}$$

Determine the constants $a$, $b$ and $n$.

My attempt to solve this problem is by trying to use the binomial theorem backwards.

The binomial theorem: $(ax + by)^{n} = \displaystyle\sum_{k = 0}^{n} {\binom{n}{k}(ax)^{n-k}(by)^{k}}$

And so if you compare the expressions, you get:

$$\binom{n}{k}(a)^{n-k}(b)^{k} = -15120$$

$n - k = 4$ and $k = 3$ so that $n = 7$

Here is where I get stuck, because now I have:

$$\binom{7}{3}(a)^{4}(b)^{3} = -15120$$

Two unknowns... How to solve it? Am I even doing it correctly?

Rodrigo de Azevedo
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Gunilla Eksjo
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    Do you really mean $(ax+ny)^n=-15120x^4y^3$? If so all below that is wrong. Or did you mean that $-15120x^4y^3$ is the term in $x^4y^3$ of the binomial expansion of $(ax+by)^n$? The the analysis below that is good. – André Nicolas May 15 '15 at 14:28
  • And you are right about non-uniqueness, $a$ and $b$ are not determined. However, if we assume they are integers, then they are completely determined. – André Nicolas May 15 '15 at 14:33
  • In the case that it is Andre's second interpretation, that you are trying to find the values of $a,b,n$ such that $-15120x^4y^3$ is one of many terms on the right hand side, and if $a$ and $b$ could be any real numbers (including irrational numbers), then there are multiple solutions. An easy solution would be $a=\sqrt[4]{15120}/\binom{7}{3}$ and $b=-1$. If you wish to remain only in the integers, then note that $(a)^4(b)^3=\frac{-15120}{\binom{7}{3}} = -432 = (-1)^{d}(2)^4(3)^3$ where $d$ is any odd integer (for example 1 or 3 or 9, etc...) – JMoravitz May 15 '15 at 14:33

3 Answers3

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I think, that the exercise is

You are given the following expression:

$(ax + by)^{n} = \color{blue}{\ldots} -15120x^{4}y^{3}+ \color{blue}{\ldots} $

Determine the constants $a$, $b$ and $n$.

In this case you know, that the 5th summand is ${7 \choose 4} \cdot (ax)^4\cdot (by)^3=-15120x^{4}y^{3}$

$15120=2^4\cdot3^3\cdot5\cdot7$.

${7 \choose 4}=5\cdot 7$

Thus $b=-3$ and $a=2$

callculus42
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The problem with your approach is that you have equated the coefficients of the $x^4y^3$ term on both sides, as if this is an equation of polynomials, but you have ignored the remaining terms.

But this cannot be an equation of polynomials to begin with, because most terms on the RHS are missing which would need to be there. For example $a^nx^n$ and $b^ny^n$.

So are we to assume $x$ and $y$ are some fixed constants?

Gregory Grant
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You get $b= (\frac{-15120}{35 \, a^{4}})^{\frac{1}{3}}$ , $a$ can be any value as we only have one term.

Leucippus
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Bob
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