SOLUTION: find the complex zeros of the polynomial function f(x)=x^4+23x^2+22 write f in factored form f(x)= please show step by step work i am very confused.

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Question 964458: find the complex zeros of the polynomial function f(x)=x^4+23x^2+22
write f in factored form f(x)=
please show step by step work i am very confused.

Found 2 solutions by rothauserc, josmiceli:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = x^4 +23x^2 +22
first write down the factors of 22, they are
1, 22
2, 11
now 1 and 22 look promising since the coefficient of x^2 is 23
let's try
(x^2 +1) * (x^2 +22) = x^4 +23x^2 + 22
we can now write the solutions for x as
x^2 = -1
x = sqrt(-1) = i
x^2 = -22
x = sqrt(-22) = i*sqrt(22)
therefore our 2 solutions for x are
x = i and x = i*sqrt(22)


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
here's how I would do it:
+f%28x%29+=+x%5E4+%2B+23x%5E2+%2B+22+
Let +z+=+x%5E2+
Substituting:
+f%28x%29+=+z%5E2+%2B+23z+%2B+22+
I can see that the factors for
finding roots
+%28+z+%2B+22+%29%2A%28+z+%2B+1+%29+=+0+
The solutions for +Z+ are:
+z+=+-22+
+z+=+-1+
------------------
So, now I can say:
+x%5E2+=+-22+
+x%5E2+=+-1+
-----------------
+x+=+sqrt%2822%29%2Ai+
+x+=+-sqrt%2822%29%2Ai+
+x+=+i+
+x+=+-i+
-------------
The 4 factors of +f%28x%29+ are:

------------------------------------------------------
check:
+%28+x+%2B+sqrt%2822%29%2Ai+%29%2A%28+x+-+sqrt%2822%29%2Ai+%29+
+x%5E2+%2B+22+
and
+%28+x+%2B+i+%29%2A%28+x+-+i+%29+=+x%5E2+%2B+1+%29+
and
+f%28x%29+=+%28+x%5E2+%2B+22+%29%2A%28+x%5E2+%2B+1+%29+
+f%28x%29+=+x%5E4+%2B+22x%5E2+%2B+x%5E2+%2B+22+
+f%28x%29+=+x%5E4+%2B+23x%5E2+%2B+22+
OK