Find the unique polynomial with real coefficients that meets these conditions----- Degree 4, zeros at x=3, x=-4, x=2-i; f(0)= -120

Question

find f(x)

A similar problem but with different degrees and zeros answer was: 2x⁴- 8x³+12x²-8x-30

Answer 1

Hi Jasmine T.,

The zero's are x = 3, x = -4, x = 2 ± i, (imaginary's come in pairs).

For x = 3, subtract 3 from both sides to get x - 3 = 0, same for the rest, x + 4 = 0, x - 2 - i = 0, and x - 2 + i = 0.

Multiply the zero's to get a polynomial:

(we will use f(x) = a(x - 3)(x + 4)(x - 2 - i)(x - 2 + i) to get our desired polynomial)

(x - 3)(x + 4)(x - 2 - i)(x - 2 + i)

(x² + 4x - 3x - 12)(x² - 2x + xi - 2x + 4 - 2i - xi + 2i - i²)

(x² + x - 12)(x² - 4x + 5)

x⁴ - 4x³ + 5x² + x³ - 4x² + 5x - 12x² + 48x - 60

x⁴ - 3x³ - 11x² + 53x - 60

So f(x) = a(x⁴ - 3x³ - 11x² + 53x - 60), now use f(0) = -120 to find a, (x = 0):

-120 = a(-60)

a = 2

Replace a into f(x):

f(x) = 2x⁴ - 6x³ - 22x² + 106x - 120

I hope this helps, Joe.

Answer 2

f(x) = a(x - 3)(x + 4)(x - (2 - i))(x - (2 + i))

-120 = a(0 - 3)(0 + 4)(0 - (2 - i))(0 - (2 + i))

Can you solve for a and answer?