Mathematics a level help!!!

how do I expand something like (x^2+3x-11)^5

its something I've found repeatedly difficult on the Mathematics admissions tests I have done so far

many thanks

Square $x^2+3x-11$.

Square this result.

Multiply this new result by $x^2+3x-11$. Job done.

how do I expand something like (x^2+3x-11)^5

its something I've found repeatedly difficult on the Mathematics admissions tests I have done so far

many thanks

You could try writing it like this



and then use binomial expansion. This may or may not be a quicker method for you - give it a go to find out.

thanks for the reply, but for certain questions I have a small time margin... is there a method you know of that could expand it somewhat faster ?

how do I expand something like (x^2+3x-11)^5

its something I've found repeatedly difficult on the Mathematics admissions tests I have done so far

many thanks

The above approach I posted is an elementary way to get there. There isn't much you can use that is elementary and *quick*.

For something slightly more advanced, you can use the trinomial expansion. Recall that for a binomial expansion

$(a+b)^n = \displaystyle \sum_{r=0}^n \binom{n}{r}a^rb^{n-r} = \sum_{r=0}^n \dfrac{n!}{r!(n-r)!}a^r b^{n-r}$

and for a trinomial we have

$(a+b+c)^n = \displaystyle \sum_{r=0}^n \sum_{s=0}^{n-r} \dfrac{n!}{r!s!(n-r-s)!}a^rb^sc^{n-r-s}$

so for something like $(x^2+3x-11)^5$ we would do

$r=0$ :

$\displaystyle \sum_{s=0}^5 \dfrac{5!}{s!(5-s)!}(3x)^s(-11)^{5-s}$
which is just the binomial expansion of $\displaystyle (3x-11)^5 = \binom{5}{0}(-11)^5 + \binom{5}{1}(3x)(-11)^4 + \ldots + \binom{5}{5}(3x)^5$.

For $r=1$ :

$\displaystyle \sum_{s=0}^4 \dfrac{5!}{1!s!(4-s)!}(x^2)(3x)^s(-11)^{4-s} = 5x^2 \left[ \binom{4}{0}(-11)^4 + \binom{4}{1}(3x)(-11)^3 + \ldots + \binom{4}{4}(3x)^4 \right]$.

and so on up to

$r=5$ :

$\displaystyle \sum_{s=0}^0 \dfrac{5!}{5!s!(0-s)!}(x^2)^5(3x)^s(-11)^{0-s} = x^{10}$

Then add up all of these results for the overall answer. A more well-laid out approach would help you do this fairly quickly with less error, but whether it's quicker for you then it's up to you to try it and see.

For instance, there is a pattern that follows here:

$\displaystyle \begin{aligned} ^5C_0 (x^2)^0 & \left[ \binom{5}{0}(-11)^5 + \binom{5}{1}(3x)(-11)^4 + \binom{5}{2}(3x)^2(-11)^3 + \binom{5}{3}(3x)^3(-11)^2 + \binom{5}{4}(3x)^4(-11) + \binom{5}{5}(3x)^5 \right] \\ + ^5C_1 (x^2) & \left[ \binom{4}{0}(-11)^4 + \binom{4}{1}(3x)(-11)^3 + \binom{4}{2}(3x)^2(-11)^2 + \binom{4}{3}(3x)^3(-11) + \binom{4}{4}(3x)^4 \right] \\ + ^5C_2(x^2)^2 & \left[ \binom{3}{0}(-11)^3 + \binom{3}{1}(3x)(-11)^2 + \binom{3}{2}(3x)^2(-11) + \binom{3}{3}(3x)^3 \right] \\ +^5C_3(x^2)^3 & \left[ \binom{2}{0}(-11)^2 + \binom{2}{1}(3x)(-11) + \binom{2}{2}(3x)^2 \right] \\ + ^5C_4(x^2)^4 & \left[ \binom{1}{0}(-11) + \binom{1}{1}(3x) \right] \\ + ^5C_5 (x^2)^5 & \end{aligned}$

which you can notice is just the binomial expansion of $(a+b)^5$ where $a=x^2, \quad b=3x-11$.