Question

factorise: x³+6x²+11x+6​

Answer 1

ANSWER:-

Considering x = (-1),

$\tt {( - 1)}^{3} + 6 {( - 1)}^{2} + 11( - 1) + 6 = 0$

$\tt = - 1 + 6 - 11 + 6 = 0$

$\sf\bold\red \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 12 - 12 = 0$

And also,

$\tt {x - 1 = 0}$

$\tt {x = 1 + 0}$

$\sf\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x = 1$

Therefore, (x = 1) is a factor.

So by performing long division method,

(Refer the above mentioned picture).

$\tt {x}^{3} + 6 {x}^{2} + 11x + 6 = (x + 1)( {x}^{2} + 5x + 6)$

$\tt = (x + 1)( {x}^{2} + 2x + 3x + 6)$

$\tt = (x + 1) |x(x + 2) + 3(x + 2)|$

$\sf\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > (x + 1)(x + 2)(x + 3)$