Find an nth-degree polynomial function with real coefficients satisfying the given conditions.

Question

n= 3

3 and 4 i are zeros

f(-1)= 136

Answer 1

Hey! Lets solve this shall we ?

We see that the given roots are 3 and 4i. We also notice one of the roots is 4i. This means there is another root of -4i when we factor the polynomial of (x²+16), obtaining ±4i. The root x = 3 came from (x-3). So, lets figure out what the 3rd degree polynomial is !

Then,

(x²+16)(x-3) , where we apply the FOIL / Distributive Property to get this polynomial. Hence,

x³-3x²+16x-48 is our polynomial of degree 3 !

I hope this helps!

Answer 2

If 'a' is a zero of a polynomial, then x-a will be a factor of the polynomial--the factor theorem. Also, if a complex number a+bi is a zero of a polynomial with real coefficients, then a-bi (the conjugate of a+bi) will be a zero also. So the polynomial you are looking for will be the product of factors P(x)=a(x-r1)(x-r2)...(x-rN), where r1, r2, ..., rN are the N zeros of the polynomial. To determine the value of 'a' use the condition P(-1)=136.

Hint, when multiplying out the factors of P(x), it is easiest to multiply out the factors with conjugate pairs first. Note that (x-(a+bi))(x-(a-bi))=(x-a)^2+b^2.