Factorize ${(x^2-4x)}{(x^2-4x-1)}-20$

${(x^2-4x)}{(x^2-4x-1)}-20$ is an algebraic expression in terms of $x$. In this problem, we have to transform this algebraic expression in product form by factorization.

Simplify the Algebraic expression

${(x^2-4x)}{(x^2-4x-1)}-20$ $\,=\,$ ${(x^2-4x)}{((x^2-4x)-1)}-20$

Now, multiply each term in the second factor by the first factor in the first term of the algebraic expression.

$= \,\,\,$ ${(x^2-4x)}{(x^2-4x)}$ $-$ $(x^2-4x) \times 1$ $-$ $20$

$= \,\,\,$ ${(x^2-4x)}^2$ $-$ $(x^2-4x)$ $-$ $20$

Convert expression in terms of another variable

The simplified algebraic expression ${(x^2-4x)}^2$ $-$ $(x^2-4x)$ $-$ $20$ can be expressed as a quadratic polynomial. In order to factorize this algebraic expression, take $u = x^2-4x$ and convert the quadratic expression in terms of $u$.

$\implies$ ${(x^2-4x)}^2$ $-$ $(x^2-4x)$ $-$ $20$ $\,=\,$ $u^2-u-20$

Factorize the Quadratic expression

Look at the third term in the quadratic expression $u^2-u-20$, the term $20$ can be factored in three ways by factorization.

$(1) \,\,\,$ $20 = 1 \times 20$
$(2) \,\,\,$ $20 = 2 \times 10$
$(3) \,\,\,$ $20 = 4 \times 5$

Choose a right combination of factoring to express the third term as product of two quantities and second term as either sum or difference form in the same combination.

$\implies$ $u^2-u-20$ $\,=\,$ $u^2-5u+4u-5 \times 4$

$u$ is a common factor in the first two terms and $4$ is a common factor in the remaining two terms. So, take each factor common from them.

$= \,\,\,$ $u(u-5)+4(u-5)$

Now, $u-5$ is a common factor in the both terms. So, take it common from them.

$= \,\,\,$ $(u-5)(u+4)$

Actually, the algebraic expression is not in terms of $u$. So, bring this algebraic expression back to original form by replacing $u$ by $x^2-4x$.

$\therefore \,\,\,$ $(u-5)(u+4)$ $\,=\,$ $(x^2-4x-5)(x^2-4x+4)$

Therefore, the algebraic expression ${(x^2-4x)}^2$ $-$ $(x^2-4x)$ $-$ $20$ is factored as $(x^2-4x-5)(x^2-4x+4)$ by factorisation.

Factorize each Quadratic expression

$(x^2-4x-5)(x^2-4x+4)$

The first factor in the product can be factored by the factoring method and the second factor can be simplified as per square of the difference formula.

$= \,\,\,$ $(x^2-5x+x-5)$ $(x^2-2 \times x \times 2+2^2)$

$x$ is a common factor in the first two terms and $1$ is common factor in the next two terms in the first factor of the algebraic expression. The second factor can be simplified as square of the difference of two terms.

$= \,\,\,$ $(x(x-5)+1(x-5)){(x-2)}^2$

Now, $x-5$ is a common factor in the both terms of the first factor of the algebraic expression. Take $x-5$ common from them and simplify the algebraic expression in factoring form.

$\therefore \,\,\,\,\,\,$ ${(x^2-4x)}^2$ $-$ $(x^2-4x)$ $-$ $20$ $\,=\,$ $(x-5)(x+1){(x-2)}^2$

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