Find Polynomial Function

[nycmathguy]

Section 2.5
Question 44

We are given 4 zeros. This leads to a fifth degree equation. Why is this the case?

In terms of 2i, the complex conjugate must also be a zero.

We now have this:

f(x) = a(x +1)(x - 5)(x - 3)(x - 2i)(x + 2i)

f(x) = a(x^3 - 7x^2 + 7x + 15)(x^2 + 4)

I know that a = 1 but what is the reason for this value of a? Why does a = 1?

f(x) = x^5 -7x^4 + 11x^3 - 13x^2 + 28x + 60

You say?

 

[MathLover1]

perfect!

 

[nycmathguy]

perfect!

We are given 4 zeros. This leads to a fifth degree equation. Why is this the case?

 

[MathLover1]

We are given 4 zeros. This leads to a fifth degree equation. Why is this the case?

that means one of given 4 zeros is complex number and it always comes in pairs

in your case given zero 2i, means you have to have -2i too

it says zeros are:
-1, 5, 3, 2i (given 4 zeros)
actually zeros are: -1, 5, 3, 2i, -2i (actually given 5 zeros)

 

[nycmathguy]

We are given 4 zeros. This leads to a fifth degree equation. Why is this the case?

that means one of given 4 zeros is complex number and it always comes in pairs

in your case given zero 2i, means you have to have -2i too

it says zeros are:
-1, 5, 3, 2i (given 4 zeros)
actually zeros are: -1, 5, 3, 2i, -2i (actually given 5 zeros)

Perfect. Thanks.