Question

(iv) (x2 - 4x)(x2 - 4x - 1) – 20​

Answer 1

$\huge\sf\underline\red{Question}$

• (x² - 4x)(x²- 4x - 1) – 20

$\huge\sf\underline\red{Answer}$

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

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LET x²+4x BE 'P'

$p(p + 1) - 20 \\ \\ p {}^{2} - p - 20 \\ \\ p {}^{2} - 5p + 4p - 20 \\ \\ p(p - 5) + 4(p - 5) \\ \\ = ( p + 4) \: \: (p - 5)$

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NOW, PUT P AS x²-4x

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

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$\bf\underbrace\red{answer:}$-$\small\fbox\pink{(x-2)²(x+1)(x-5)}$

Answer 2

Appropriate Question :-

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

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:( {x}^{2} - 4x)( {x}^{2} - 4x - 1) - 20$

Let we assume that,

$\red{ \boxed{ \sf{ \: {x}^{2} - 4x = y}}}$

So, above expression can be rewritten as

$\rm \:  =  \: y(y - 1) - 20$

$\rm \:  =  \: {y}^{2} - y - 20$

Now, using the concept of splitting of middle terms, we have

$\rm \:  =  \: {y}^{2} -5y + 4y - 20$

$\rm \:  =  \: y(y - 5) + 4(y - 5)$

$\rm \:  =  \: \:(y - 5)(y + 4)$

$\rm \:  =  \: ( {x}^{2} - 4x - 5)( {x}^{2} - 4x + 4)$

Again, on splitting the middle terms, we get

$\rm \:  =  \: ( {x}^{2} - 5x + x - 5)( {x}^{2} - 2x - 2x + 4)$

$\rm \:  =  \: \bigg[x(x - 5) + 1(x - 5)\bigg]\bigg[x(x - 2) - 2(x - 2)\bigg]$

$\rm \:  =  \: (x - 5)(x + 1)(x - 2)(x - 2)$

Hence,

$\red{ \boxed{ \sf{ \:( {x}^{2} - 4x)( {x}^{2} - 4x - 1) - 20 =  \: (x - 5)(x + 1)(x - 2)(x - 2)}}}$

Additional Information :-

More Identities to know :-

$\red{ \boxed{ \sf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

$\red{ \boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

$\red{ \boxed{ \sf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}}$

$\red{ \boxed{ \sf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}}$

$\green{ \boxed{ \sf{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy}}}$

$\green{ \boxed{ \sf{ \: {(x + y)}^{2} = {(x - y)}^{2} + 4xy}}}$

$\green{ \boxed{ \sf{ \: {(x - y)}^{2} = {(x + y)}^{2} - 4xy}}}$

$\blue{ \boxed{ \sf{ \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}}$

$\blue{ \boxed{ \sf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}}}$

$\blue{ \boxed{ \sf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}}$

$\blue{ \boxed{ \sf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2})}}}$