Finding k so that -2-i is a zero of f(x)

[pog99]

Hey,
i have been having trouble with this problem.
It reads:
If f(x)=x^3+kx^2-7x-15, find the value of k so that -2-i is a zero of f(x)

I have already attempted to solve this equation by making -2-i a zero of f(x) like this

-2-il 1 k 7 -15

and then multiplying and adding though to find the value the remainder and then setting it equal to 0 but i came up with wierd numbers and can't get an answer

Thanks,
Paul

 

[soroban]

Hello, Paul!

This is easier than you think . . .

If \(\displaystyle f(x)\;=\;x^3\,+\,kx^2\,-\,7x\,-\,15\),
find the value of \(\displaystyle k\) so that \(\displaystyle \,-2-i\,\) is a zero of \(\displaystyle f(x)\).

Since \(\displaystyle x\:=\:-2\,-\,i\) is a zero of \(\displaystyle f(x)\), then: \(\displaystyle \:f(-2-i) \:=\:0\)


We have: \(\displaystyle \:f(-2-i)\;=\;(-2-i)^3 \,+\,k(-2-i)^2\,-\,7(-2-i)\,-\,15\;=\;0\)

. . which simplfies to: \(\displaystyle \:3(k\,-\,1)\,+\,4(k\,-\,1)i\;=\;0\)


Therefore: \(\displaystyle \:k\:=\:1\)

 

[pog99]

thanks a lot, i thought about doing it that way first but thought i was wrong

----Paul