Problem Solving with Binomial Expansion (AQA GCSE Further Maths): Revision Note

• Look out for extra information about unknowns

  • e.g. if it says "...where $p>0$" and you have $p=\pm 3$ then use the positive value

• If you forget how to find a specific term in the exam, just expand the whole binomial using Pascal's triangle then find it

(a) Find the coefficient of $x^3$ in the expansion of ${\left(4x-2\right)}^5$.
 

Imagine $a=\left(4x\right)$ and $b=\left(-2\right)$ in ${\left(a+b\right)}^5$

The row from Pascal's triangle that starts 1, 5, ... is

$\begin{array}{cccccc}1& 5& 10& 10& 5& 1\end{array}$

The term required is $x^3$, which is third along when considering $x^5,x^4,x^3,...$

$\begin{array}{cccccc}1& 5& \overline{)10}& 10& 5& 1\end{array}$

Therefore the term required is $10a^3{b}^2$ in the expansion of ${\left(a+b\right)}^5$

$10\times {\left(4x\right)}^3\times {\left(-2\right)}^2$

So the coefficient of $x^3$ is

$\begin{array}{rcl}10\times 4^3\times {\left(-2\right)}^2& =& 10\times 64\times 4\\ & =& 2560\end{array}$

The coefficient of the $x^3$ term is 2560

(b) Given that $p>0$ and that the coefficient of $x^4$ in the expansion of ${\left(3x-p\right)}^6$  is 59 535, find the value of $p$.

Imagine $a=\left(3x\right)$ and $b=\left(-p\right)$ in ${\left(a+b\right)}^6$

The row from Pascal's triangle that starts 1, 6, ... is

$\begin{array}{cccccc}1& 6& 15& 20& 15& 61\end{array}$

The term required is $x^4$, which is third along when considering $x^6,x^5,x^4,...$

$\begin{array}{cccccc}1& 6& \overline{)15}& 20& 15& 61\end{array}$

Therefore the term required is $15a^4{b}^2$ in the expansion of ${\left(a+b\right)}^6$

$15\times {\left(3x\right)}^4\times {\left(-p\right)}^2$

So the coefficient of $x^4$ is

$\begin{array}{rcl}15\times 3^4\times {\left(-p\right)}^2& =& 1215p^2\end{array}$

The question gives the coefficient as 59 535 so set up and solve an equation for $p$

$\begin{array}{rcl}1215p^2& =& 59535\\ p^2& =& 49\end{array}$

$p>0$ so the positive square root is needed

The value of $p$ is 7

(c) Find the coefficient of $x$ in the expansion of ${\left(2x+\frac{1}{x}\right)}^5$.

Imagine $a=\left(2x\right)$ and $b=\left(\frac{1}{x}\right)$ in ${\left(a+b\right)}^5$

The row from Pascal's triangle that starts 1, 5, ... is

$\begin{array}{cccccc}1& 5& 10& 10& 5& 1\end{array}$

The term required is in $x$, but this time both terms in the binomial depend on $x$ so consider how their powers are multiplied together in the expansion, ${\left(2x\right)}^5{\left(\frac{1}{x}\right)}^0,{\left(2x\right)}^4{\left(\frac{1}{x}\right)}^1,{\left(2x\right)}^3{\left(\frac{1}{x}\right)}^2,...$

The third along gives$8x^3\times \frac{1}{x^2}=8x$, so select the third number from the row in Pascal's triangle

$\begin{array}{cccccc}1& 5& \overline{)10}& 10& 5& 1\end{array}$

The term required is $10a^3{b}^2$ in the expansion of ${\left(a+b\right)}^5$

$10\times {\left(2x\right)}^3\times {\left(\frac{1}{x}\right)}^2$

So the coefficient of $x$ is

$\begin{array}{rcl}10\times 2^3& =& 80\end{array}$

The coefficient of the $x$ term is 80