 QUESTION

# Factorises the given expression: $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$.

Hint: Mathematics includes the study of topics which are related to quantity, structure, space and change. To solve this problem, we first try to simplify the expression. For doing so we let ${{x}^{2}}-4x=t$ and solve by using this consideration. In this way we can easily factorise the expression.

According to a question, we are given the expression $\left( {{x}^{2}}-4x \right)\left( {{x}^{2}}-4x-1 \right)-20$. To simplify this complex expression, we let ${{x}^{2}}-4x=t$.
This can be mathematically expressed as: $t(t-1)-20$.
Since, these types of factorization are familiar to us, so we try to factorise the equation in terms of t:
${{t}^{2}}-t-20$
Now, by using the middle term splitting method of separating the variables, we get
\begin{align} & {{t}^{2}}+4t-5t-20=0 \\ & t(t+4)-5(t+4)=0 \\ & (t-5)(t+4)=0 \\ \end{align}
So, the factors of the expression in terms of t are (t-5) and (t+4).
Now, finally replacing t as ${{x}^{2}}-4x$,we get
$\left( {{x}^{2}}-4x-5 \right)\left( {{x}^{2}}-4x+4 \right)=0$
On further simplifying this expression, by using middle term splitting method again in part one:
\begin{align} & {{x}^{2}}-4x-5={{x}^{2}}+x-5x-5 \\ & \Rightarrow x(x+1)-5(x+1)=0 \\ & (x-5)(x+1)=0 \\ \end{align}
Now, for the second part, using the identity of ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get
${{x}^{2}}-4x+4={{(x-2)}^{2}}$
$\therefore (x-5)(x+1){{(x-2)}^{2}}=0$
Hence, the factors of the required equation as per problem statement are: $(x-5)(x+1){{(x-2)}^{2}}=0$

Note: This question can be alternatively solved by expanding the powers of x in the given expression and then letting ${{x}^{2}}$ as another variable. But the complexity of the problem would increase by doing so. Hence, for simplification we must assume some other variable which can be finally replaced in the original expression to obtain the answer.