Question

Factorise: x3(3 is cube) + 6x2(2 is squre) +11x +6.​

Answer 1

Answer:

According to Factor theorem, if (x - a) is a polynomial factor f(x), then f(a) = 0

Let f(x) = x^{3}-6 x^{2}+11 x-6f(x)=x

3

−6x

2

+11x−6

Let us check if (x - 1) is the factor of f(x),

Then,

f(1) = 1^{3}-6\left(1^{2}\right)+11(1)-6=1-6+11-6=0f(1)=1

3

−6(1

2

)+11(1)−6=1−6+11−6=0

Therefore (x-1) is a factor of f(x)

Let us check for the other factors

Hence,

f(x)=(x-1)\left(x^{2}-5 x+6\right)f(x)=(x−1)(x

2

−5x+6)

x^{2}-5 x+6=x^{2}-2 x-3 x+6x

2

−5x+6=x

2

−2x−3x+6

=x(x-2)-3(x-2)=x(x−2)−3(x−2)

= (x - 2)(x - 3)=(x−2)(x−3)

f(x) = (x - 1)(x - 2)(x - 3)f(x)=(x−1)(x−2)(x−3)

Therefore, 1, 2, 3 are the factors of f(x)