Question

Factorize
(x+2)(x+3)x+4)(x+5)-15​

Answer 1

Required Answer:-

Given to factorise:

• (x + 2)(x + 3)(x + 4)(x + 5) - 15

Solution:

We have,

(x + 2)(x + 3)(x + 4)(x + 5) - 15

= (x + 2)(x + 5)(x + 3)(x + 4) - 15

= [x(x + 5) + 2(x + 5)] × [x(x + 4) + 3(x + 4)] - 15

= (x² + 5x + 2x + 10)(x² + 3x + 4x + 12) - 15

= (x² + 7x + 10)(x² + 7x + 12) - 15

= (y + 10)(y + 12) - 15 [Taking y = x² + 7x]

= y(y + 12) + 10(y + 12) - 15

= y² + 12y + 10y + 120 - 15

= y² + 22y + 105

Now, find two numbers whose sum is 22 and product is 105.

• 15 × 7 = 105
• 15 + 7 = 22

So, it will be equal to,

= y² + 15y + 7y + 105

= y(y + 15) + 7(y + 15)

= (y + 7)(y + 15)

Substitute back y = x² + 7x,

= (x² + 7x + 7)(x² + 7x + 15)

Now, this can't be factorised more.

★ Hence, the factorised form is (x² + 7x + 7)(x² + 7x + 15)

Answer:

• The factorised form is (x² + 7x + 7)(x² + 7x + 15)