Question

Given that -2 + i is a zero of
f(x) = x^4 + 8x^3 + 23x^2 + 28x + 10
Find all other zeros and write the complete factorization.

Answer 1

if -2+i is a zero, -2-i must also be a solution. so this is part of the factorization.\[\Large (x-(-2+i)) (x-(-2-i)\]
You might want to expand that out, then divide the original f(x) by it, then find the zeroes of the quotient quadratic.

Answer 2

How would you divide the original function by it? Since it has the imaginary number in it?

Answer 3

Expand that out first, the i's should disappear.

Answer 4

expand how?
\[(\chi+2-i)(\chi+2+i)\]
?

Answer 5

Expand... as in multiply it out, distribute.

Answer 6

The result will have no i's, then you can divide f(x) by it.

Answer 7

\[given x=-2+\iota is a zero of the eq\]
\[\therefore x=-2-\iota is also a zero of the solution\]
\[\left( x+2-\iota \right) and \left( x+2+\iota \right) are factors of given eq.\]
\[\left( x+2-\iota \right)\left(x+2+\iota \right)=\left( x+2 \right)^{2}-\iota ^{2}=x ^{2}+4x+4+1=x ^{2}+4x+5\]

\[x ^{4}+8x ^{3}+23x ^{2}+28x+10= \left(x ^{2}+4x+5 \right)\left(x ^{2}+4x+2 \right)\]
\[x ^{2}+4x+2=(x ^{2}+4x+4)-2=\left( x+2 \right)^{2}-\left( \sqrt{2} \right)^{2}\]
\[=\left( x+2+\sqrt{2} \right)\left( x+2-\sqrt{2} \right)\]
\[x ^{4}+8x ^{3}+23x ^{2}+28x+10=\[\left( x+2-\iota \right)\left( x+2+\iota \right)\left( x+2+\sqrt{2} \right)\left( x+2-\sqrt{2} \right)\]
\[zeros are -2-\iota,-2+\iota,2+\sqrt{2},2-\sqrt{2}\]