182k views
5 votes
Third-degree, with zeros of 4-I,4+I and a leading coefficient of -5

User Chardy
by
8.4k points

1 Answer

5 votes
Please use "i" for the imaginary operator, not "I."

A third degree poly would have 3 factors and 3 zeros; you've supplied only 2 zeros.

Let's model the polynomial as y = ax^3 + bx^2 + cx + d.
We are told that the leading coefficient is -5. Therefore, our model becomes

y = -5x^3 + bx^2 + cx + d

The two given roots are 4-i and 4+i; if either is subst. into the above equation, y takes on the value 0 (this is the meaning of "root")

Thus, if x = 4-i, 0 = -5(4-i)^3 + b(4-i)^2 + c(4-i) + d.
If x = 4+i, 0 = -5(4+i)^3 + b(4+i)^2 + c(4+i) + d

Here we have two equations in three unknowns (b, c and d). There's not enough info to enable us to determine all three b, c and d.

One thing you might do at this point would be to set d=0 arbitrarily, and then solve the resulting two equations (without d) for b and c.

Can you take the ball and run with it from this point on?
User Krupan
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.