Question

P(x) = 207 +9x² +6012-111-6, g(x)=x-9​

Answer 1

Step-by-step explanation:

The zeroes are -2, 10 and 1.

Step-by-step explanation:

Given, polynomial,

p(x)=x^3-9x^2-12x+20p(x)=x

3

−9x

2

−12x+20

Since, (x+2) is a factor of p(x).

Thus, p(x) must be divisible by (x+2),

When we divide p(x) by x + 2,

The quotient is,

x^2-11x+10x

2

−11x+10

Hence, we can write,

x^3-9x^2-12x+20=(x+2)(x^2-11x+10)x

3

−9x

2

−12x+20=(x+2)(x

2

−11x+10)

For zeroes,

p(x) = 0,

\implies (x+2)(x^2-11x+10)=0⟹(x+2)(x

2

−11x+10)=0

(x+2)(x^2-10x-x+10)=0(x+2)(x

2

−10x−x+10)=0

(x+2)(x(x-10)-1(x-10)=0(x+2)(x(x−10)−1(x−10)=0

(x+2)(x-10)(x-1)=0(x+2)(x−10)(x−1)=0

Thus, zeroes of p(x) are -2, 10 and 1.

Answer 2

Step-by-step explanation:

I hope it's helpful to you