Question

factorise (x^2 +7x)^2 +22(x^2 +7x) +105

Answer 1

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

Given that

$\rm :\longmapsto\: {( {x}^{2} + 7x) }^{2} + 22( {x}^{2} + 7x) + 105$

To factorize this expression, let assume that

$\red{\rm :\longmapsto\: {x}^{2} + 7x = y}$

So, above expression reduced to

$\rm :\longmapsto\: {y}^{2} + 22y + 105$

Now, it's a quadratic polynomial can be factorized with the help of splitting of middle terms.

We have to split 105 in such a way that on multiplication, we get 105 and on addition we get 22.

So, required splitting is 15 and 7

So,

$\rm \:  =  \:  \: \: {y}^{2} + 15y + 7y + 105$

$\rm \:  =  \:  \: \:y(y + 15) + 7(y + 15)$

$\rm \:  =  \:  \: \:(y + 7)(y + 15)$

On Substituting the value of y, we get

$\rm \:  =  \:  \: \:( {x}^{2} + 7x + 7)( {x}^{2} + 7x + 15)$

Hence,

$\bf :\longmapsto\: {( {x}^{2} + 7x) }^{2} + 22( {x}^{2} + 7x) + 105$

$\bf\:  =  \:  \: \:( {x}^{2} + 7x + 7)( {x}^{2} + 7x + 15)$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)