Question

${x}^{3} + {6 {x}^{2} } + {11}x + 6$
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Answer 1

Answer:

Step-by-step explanatio

Possible values of x can be factors of 6 that +-1,+-2,+-3,+-6

Putting x =1

1^3+6(1)^2+11(1)+6

24≠0

Putting x = -1

-1^3+6(-1)^2+11(-1)+6

-1+6-11+6

12-12

0

So x=-1 is a zero of p(x)

x+1=0

Dividing whole equation with x+1 you will get

x^2+5x+6

x^2+3x+2x+6

x(x+3)+2(x+3)

(x+3)(x+2)

So the factors are

(x+1)(x+3)(x+2) ans