Question

Using Factor Theorem, factorise the following polynomial:
x^3+6x^2+11x+6. ​

Answer 1

hii

Step-by-step explanation:

Zeroes of x

3

−6x

2

+11x−6

By using rational theorem, the roots can be among the factors of

1

6

=6

Let us try x=1

⇒(1)

3

−6(1)

2

+11(1)−6=0

∴(x−1) is a factor of x

3

−6x

2

+11x−6.

Now, using synthetic division method :

So, the quotient =x

2

−5x+6

Now, using common factor theorem,

⇒x

2

−5x+6=x

2

−2x−3x+6

=x(x−2)−3(x−2)

=(x−2)(x−3)

∴ Zeroes of the polynomial =1,2,3

So, factor of the polynomial =(x−1)(x−2)(x−3).

Hence, the answer is (x−1)(x−2)(x−3).

Answer 2

HOLA MATE!!!!!!

Let f (x) = x3+6x2+11x+6 be the given polynomial.

The constant term in f (x) is 6 and factors of 6 are

Putting x = - 1 in f (x) we have,

f (-1) = (-1)3 + 6 (-1)2 + 11 (-1) + 6

= -1 + 6 – 11 + 6

= 0

Therefore, (x + 1) is a factor of f (x)

Similarly, (x + 2) and (x + 3) are factors of f (x).

Therefore, f (x) = k (x + 1) (x + 2) (x + 3)

x3+6x2+11x+6 = k (x + 1) (x + 2) (x + 3)

Putting x = 0, on both sides we get,

0 + 0 + 0 + 6 = k (0 + 1) (0 + 2) (0 + 3)

6 = 6k

k = 1

Putting k = 1 in f (x) = k (x + 1) (x + 2) (x + 3), we get

f (x) = (x + 1) (x + 2) (x + 3)

Hence,

x3+6x2+11x+6 = (x + 1) (x + 2) (x + 3)