# Thread: What does this mean in my exam practice paper?

1. ## What does this mean in my exam practice paper?

Find the independant of x in the expansion of [x-(2/x)]^4

Thanks, Rob

2. That means find the constant term, the term without x. In binomial expansions, this is usually the term with $x^0$ (x to the power zero)

3. I'm not sure that I quite understand that. Surely with the binomial expansion shown above, all of the expanded terms world have an x somewhere.

4. Note: $\left(x-\frac{2}{x}\right)^4$ can be written as $(x-2(x^{-1}))^4$

Just looking at the x terms (ignoring coefficients):

$(x^4)(x^0) + (x^3)(x^{-1}) + (x^2)(x^{-2}) + (x^1)(x^{-3}) + (x^0)(x^{-4})$

The third term cancels out.

5. Hello, Rob!

Don't argue about it . . . just expand it!

Find the constant term in the expansion of . $\left(x-\frac{2}{x}\right)^4$

$\left(x-\frac{2}{x}\right)^4 \;\;=\;\;(x)^4 - \:4(x)^3\!\left(\frac{2}{x}\right) + \:6(x)^2\!\left(\frac{2}{x}\right)^2 - \:4(x)^1\!\left(\frac{2}{x}\right)^3 + \left(\frac{2}{x}\right)^4$

. . . . . . . $= \;\;x^4 \:- \;4x^3\!\cdot\!\frac{2}{x} \:+ \;6x^2\!\cdot\!\frac{4}{x^2} \:- \;4x\!\cdot\!\frac{8}{x^3} \:+ \;\frac{16}{x^4}$

. . . . . . . $= \;\;x^4 \:-\: 8x^2 \:+\: \underbrace{{\color{blue}24}}_{\uparrow} \:-\: \frac{32}{x^2} \:+\: \frac{16}{x^4}$