Find $k$ if given the constant term of a binomial expression?

Question

Consider the expansion of $x^2(3x^2+\frac{k}{x})^8$. The constant term is $16,128$. Find $k$.

This is simply an example of a type of question I cannot understand how to do. I have many questions:

1) What is a constant term in regards to binomials?

2) How would I go about solving this?

3) How can I check that my solution is correct?

4) How would I be able to find the constant term if asked to do so?

Thanks very much :)

Answer 1

The constant term is just the coefficient of $x^0$; it's just like the constant term of a polynomial. So to find the constant term, you want to figure out what is the coefficient of the term in $(3x^2 + \frac{k}{x})^8$ corresponding to $x^{-2}$, since this will cancel the $x^2$ to produce a constant. To do that, you can expand $(3x^2 + \frac{k}{x})^8$ using the binomial theorem.

Answer 2

The constant term in $x^2(3x^2+\frac{k}{x})^8$ is the term in the expansion that does not have any $x$ factor. Let's use the binomial theorem to expand this out to find the term without the $x$ factor.

$$ \begin{aligned} x^{2}\sum_{i=0}^{8}\binom{8}{i}(3x^2)^{i}(\frac{k}{x})^{8-i}&=\sum_{i=0}^{8}\binom{8}{i}3^{i}k^{8-i}\frac{x^{2i}x^{2}}{x^{8-i}}\\ &=\sum_{i=0}^{8}\binom{8}{i}3^{i}k^{8-i}x^{2i+2-8+i}\\ &=\sum_{i=0}^{8}\binom{8}{i}3^{i}k^{8-i}x^{3i-6} \end{aligned} $$

Now, $x^{3i-6}=x^{0}\iff i = 2$. This tells us that the constant term is:

$$ \binom{8}{2}3^{2}k^{6}=\frac{8!}{2!6!}\cdot 9k^{6}=28\cdot 9 k^{6}=252k^6. $$

Now, we solve for $k$:

$$ 252k^6=16128\implies k^6=64\implies k=2 , -2 $$

Answer 3

A good way of getting your head round the problem is to rewrite it.

You are being asked about the coefficient of $x^0$ in the binomial expansion of $x^2(3x^2+\frac{k}{x})^8$.

Multiply the expression inside the parentheses by $x$ and divide the expression outside the parentheses by $x^8$ (the 8th power is because the expression inside the parentheses is being raised to the 8th power):

$$x^2(3x^2+\frac{k}{x})^8 \equiv \frac{x^2}{x^8}(3x^3+k)^8$$

So you are being asked about the coefficient of $x^0$ in $x^{-6}(3x^3+k)^8$.

Or, equivalently, about the coefficient of $x^6$ in $(3x^3+k)^8$.

At which point, this should look like the kind of binomial problem you are familiar with. I don't want to spoil it for you by giving the answer.