Question

Q8.) Determine whether the input values are correct. Be sure to example each answer (why it's correct, or wrong).

Answer 1

f(x) = x^3 + 8x^2 - 4x - 32

= (x^3 + 8x^2) - (4x + 32)

= x^2(x + 8) - 4(x + 8)

= (x^2 - 4)(x + 8)

= (x -2)(x + 2)(x + 8)

f has zeros at x=2, -2, -8

with multiplicity of 1 for all three zeros.

Answer 2

A "zero" of a polynomial is a value of xx at which the polynomial = 0

'multiplicity' is that the zero is actually a solution "multiple times"

f(x) = x^3 + 8x^2 - 4x - 32

f(2) = 2^3 + 2x^2 - 2*2 - 32 = 0

f(-2) = -2^3 + 8(-2)^2 - 4(-2) - 32 = 0

f(-8) = (-8)^3 + 8(-8)^2 - 4(-8) - 32 = 0

so 2, -2 and -8 are the three zeros

f(x) = (x -2)(x +2)(x+8)

multiplicity of all 3 zeros is 1