SOLUTION: Solve: (x^4)-(8x^3)+(19x^2)-(16x)+(34)=0... Given that 4-i is a root Show your work please. I am completely lost. O

SOLUTION: Solve: (x^4)-(8x^3)+(19x^2)-(16x)+(34)=0... Given that 4-i is a root Show your work please. I am completely lost. O_o

Question 172463: Solve: (x^4)-(8x^3)+(19x^2)-(16x)+(34)=0... Given that 4-i is a root
Show your work please. I am completely lost. O_o
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Complex solutions to polynomial always come in conjugate pairs.
So, if 4-i is a solution, then so is 4+i.
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I usually graph functions to try get a better sense of where the zeros are.
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As you see here, the function does get close to crossing the x axis which means all of the roots are complex.
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Since we know that (4-i) and (4+i) are both roots, then

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We can divide (using polynomial long division) the original polynomial with this polynomial to find the remainder and then use the quadratic formula on the remainder to find its roots.
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First multiplier:
Remainder:
Second multiplier:
Remainder:
The other polynomial is

The roots are:

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So the 4 roots of are

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