Question

Which of the following is a polynomial function in factored form with zeros at –6, –2, and 3? a.f(x) = (x + 6)(x + 2)(x – 3) b.f(x) = x3 + 5x2 – 12x – 36 c.f(x) = (x – 6)(x – 2)(x + 3) d.f(x) = x3 – 5x2 – 12x + 36

Answer 1

Answer:

a) f(x) = (x + 6)(x + 2)(x – 3)

b) $f(x) = x^3 + 5x^2 - 12x - 36$

Step-by-step explanation:

We have to find the polynomial with zeroes as -6, -2, and 3.

Roots are those values of x for which the polynomial is zero.

a) f(x) = (x + 6)(x + 2)(x – 3)

$f(x) = (x + 6)(x + 2)(x -3)=0\\\Rightarrow x+6 = 0, x+2 = 0, x-3 = 0\\\Rightarrow x = -6,x = -2, x = 3$

b)

$f(x) = x^3 + 5x^2 - 12x - 36 \\\text{It can be factored as}\\f(x) = (x+6)(x-3)(x+2) = 0\\\Rightarrow (x+6)=0,(x-3)=0,(x+2)=0\\\Rightarrow x = -6, x = 3, x = -2$

c) f(x) = (x – 6)(x – 2)(x + 3)

$f(x) = (x - 6)(x - 2)(x + 3) =0\\\Rightarrow (x -6)(x -2)(x + 3) = 0\\\Rightarrow (x -6)=0,(x -2)=0,(x + 3)=0\\\Rightarrow x = 6, x = 2, x = -3$

d)

$f(x) = x^3 - 5x^2 - 12x + 36\\f(x) = (x-2)(x+3)(x-6)=0\\\Rightarrow (x-2) = 0, (x+3)=0,(x-6)=0\\\Rightarrow x =2, x = -3, x = 6$