Question

IF -2 IS A ZERO OF THE CUBIC POLYNOMIALS P(X)=X^3+6X^2+11X+6 FIND THE REMAINING ZEROS OF P(X).​

Answer 1

Answer:

Other two zeros are -3 and -1

Explanation:

Given polynomial,

⇒ p(x) = x³ + 6x² + 11x + 6

Whose one of the zeros is -2

We need to find the another two zeros

Since, -2 is a zero then, we can say that (x + 2) is a factor of p(x)

On dividing p(x) by its factor we get the quotient as another factor.

x + 2)x³ + 6x² + 11x + 6(x² + 4x + 3

(-)

x³ + 2x²

4x² + 11x

4x² + 8x

3x + 6

3x + 6

0

∴ Another factor we got is x² + 4x + 3

By middle term splitting,

→ x² + 3x + x + 3

→ x(x + 3) + 1(x + 3)

→ (x + 3)(x + 1)

Then,

⇒ x + 3 = 0

⇒ x = -3

And also,

⇒ x + 1 = 0

⇒ x = -1

Hence, the other zeros are :- -3 and -1

Answer 2

Step-by-step explanation:

Answer:

Other two zeros are -3 and -1

Explanation:

Given polynomial,

→ p(x) = x³ + 6x² + 11x + 6

Whose one of the zeros is -2

We need to find the another two zeros

Since, -2 is a zero then, we can say that

(x + 2) is a factor of p(x)

On dividing p(x) by its factor we get the quotient as another factor.

x + 2)x³ + 6x² + 11x + 6(x² + 4x +3

(-)

x³ + 2x²

4x² + 11x

4x² + 8x

3x + 6

3x + 6

.. Another factor we got is x² + 4x + 3

By middle term splitting,

x² + 3x + x + 3

→x(x+3)+1(x+3)

(x+3)(x + 1)

:. Another factor we got is x² + 4x + 3

By middle term splitting,

x² + 3x + x + 3

x(x+3)+1(x + 3)

- (x+3)(x + 1)

Then,

x+3=0

⇒ x = -3

And also,

⇒ x+1=0

⇒ x = -1