Create a degree 3 polynomial in the form f(x)=ax3+bx2+cx+d such that f(-4)=0 and f(-3-i)=0

Answer 1

Use the factored form for a degree three polynomial (three factors):

$f(x)=a(x-p)(x-q)(x-r)$

where p, q, and r are the zeros. We already know that -4 and -3-i are zeros. For real coefficients, if -3-i is one zero, so is its conjugate -3+1. Hence p = -4, q = -3-i, r = -3+i. a is a constant. Since we are free to create any degree 3 polynomial that has the specified factors, we can choose a = 1:

$f(x)=(x-(-4))(x-(-3-i))(x-(-3+i))$

$f(x)=(x+4)(x+3+i)(x+3-i)$

Multiply it out to put it into standard form (f(x) = ax3 + bx2 + cx +d)

Answer 2

This question was posted before and correctly answered. Here is the answer again:

It must have have 3 zeros. For the coefficients to be real, the complex conjugate of a complex zero must also be a zero. Hence, the 3 zeros are

 

-4

-3 - i

-3 + i

 

f(x) = A(x + 4) [x - (-3 - i)] [x - (-3 + i)]

 

where A is real and A ≠ 0. You can expand to standard form and finish from here.