Find a polynomial function of degree 3 with real coefficients that has the given zeros of -3, -1 and 4 for which f(-2) = 24.

Solution:

Given, the roots of the polynomial function of degree 3 are -3, -1 and 4.

f(-2) = 24

So, f(x) = a(x - 4)(x + 3)(x + 1) 

f(x) = a(x² + 3x - 4x - 12) (x + 1)

= a(x² - x - 12)(x + 1)

= a(x³ + x² - x² - x - 12x - 12)

f(x) = a(x³ - 13x - 12)

f(-2) = 24 = a(x³ - 13x - 12)

a(((-2)³ - 13(-2) - 12) = 24

a(-8 + 26 - 12) = 24

a(26 - 20) = 24

a(6) = 24

a = 24/6

a = 4

Therefore, polynomial function is 4(x - 4)(x + 3)(x + 1) 

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Find a polynomial function of degree 3 with real coefficients that has the given zeros of -3, -1 and 4 for which f(-2) = 24.

Summary:

A polynomial function of degree 3 with real coefficients that has the given zeros of -3, -1 and 4 for which f(-2) = 24 is 4(x - 4)(x + 3)(x + 1).