Give a third degree polynomial that has zeros of 3, 3i, and −3i, and has a value of −13 when x=2. Write the polynomial in standard form

Question

The 3 zeros are given but how is this expressed in standard form? And how can I prove that the function has a value of -13?

Answer 1

P(x) = k(x-3) (x-3i)(x+3i)

= k (x-3)(x^2 + 9)

= k(x^3 - 3x^2 + 9x - 27)

P(2)=-13 as given

k [ (2)^3 - 3(2)^2 + 9(2) - 27 ] = -13

k [ 8 - 3(4) + 9(2) - 27] = -13

k [ 8 - 12 + 18 - 27] = -13

k [ -4+18-27] = -13

k [ 14 - 27] = -13

k [ -13] = -13

k = 1

P(x) = (x-3)(x^2 + 9)

= (x^3 - 3x^2 + 9x - 27)

Answer 2

Put the polynomial into factored form first. Since there are three zeros, there are three factors:

P(x) = a·(x-3)·(x-3i)·(x+3i)

where a is a constant known as the leading coefficient. To find the value of a, plug in the given point, (2, -13):

P(x) = a·(x-3)·(x-3i)·(x+3i)

-13 = a·(2-3)·(2-3i)·(2+3i)

Multiply it out and solve for a, then plug that value of a into the P(x) formula above. You then have the complete polynomial in factored form. To put it into standard form (P(x) = ax³ + bx² + cx + d), multiply P(x) out and arrange it in descending powers of x.

Answer 3

(x-3)(x^2+9) =

x^3 -3x^2 +9x -27