Algebraic Factorisation with Exponents (Indices)

Algebraic Factorisation with Exponents (Indices)

$\textit{Factorisation}$

We first look for $\textit{common factors}$ and then for other forms such as $\textit{perfect squares}$, $\textit{difference of two squares}$, etc.

Example 1

Factorise $2^{n+4} + 2^{n+1}$.

\( \begin{align} \displaystyle
&= 2^{n+1} \times 2^{3} + 2^{n+1} \\
&= 2^{n+1}(2^{3} + 1) \\
&= 2^{n+1} \times 9
\end{align} \)

Example 2

Factorise $2^{n+3} + 16$.

\( \begin{align} \displaystyle
&= 2^{n+3} + 2^4 \\
&= 2^4 \times 2^{n-1} + 2^4 \\
&= 2^4 (2^{n-1} + 1) \\
&= 16 (2^{n-1} + 1)
\end{align} \)

We can use the difference of squares rule:
$$ \large a^2-b^2 = (a-b)(a+b)$$

Example 3

Factorise $9^x-16$.

\( \begin{align} \displaystyle
&= (3^2)^x-4^2 \\
&= (3^x)^2-4^2 \\
&= (3^x-4)(3^x+4)
\end{align} \)

We can use the perfect square rules:
$$ \large \begin{align} a^2 + 2ab + b^2 &= (a+b)^2 \\
a^2-2ab + b^2 &= (a-b)^2 \end{align} $$

Example 4

Factorise $9^x + 4 \times 3^x + 4$.

\( \begin{align} \displaystyle
&= (3^2)^x + 4 \times 3^x + 4 \\
&= (3^x)^2 + 4 \times 3^x + 4 \\
&= (3^x+2)^2
\end{align} \)

Example 5

Factorise $4^x-2^x-6$.

\( \begin{align} \displaystyle
&= (2^2)^x-2^x-6 \\
&= (2^x)^2-2^x-6 \\
&= (2^x + 2)(2^x-3)
\end{align} \)

Example 6

Factorise $3^{2x+1} + 6 \times 3^x-24$.

\( \begin{align} \displaystyle
&= 3 \times 3^{2x} + 6 \times 3^x-24 \\
&= 3 (3^{2x} + 2 \times 3^x-8) \\
&= 3 \left[(3^x)^2 + 2 \times 3^x-8\right] \\
&= 3(3^x-2)(3^x + 4)
\end{align} \)

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