How many complex zeros does the polynomial function have?
f(x)=x^4+5x2+6
A.2
B.3
C.4
D.5
Find all 4 factors of the polynomial function over the complex numbers.
f(x)=x^4+3x^2−4
Graph it to find the two real zeros.
Depress the poly with each zero.
Solve the remaining quadratic polynomial to reveal the remaining roots.
A.(x+1)(x−1)(x+2i)(x−2i)
B.(x+i)(x−i)(x+4i)(x−4i)
C.(x+1)(x−1)(x+4)(x−4)
D.(x+2i)(x−2i)
+895 1) For the first, the number of complex zeros you have is always the same as your highest degree.
So, for that one it would be 4 complex zeros.
2) f(x)=x4+3x2−4
Here is the graph: https://www.desmos.com/calculator/zm7f4iwvrb
From the graph, the two zeros are (-1,0) and (1,0).
Now, I'm going to use long division for these.
Pull off all the coefficients (Don't forget the missing terms. This problem is really x4+0x3+3x2+0x-4)
I'm gong to use (-1,0) first, since it makes x+1 as a factor.
(Sorry if the picture is a bit fuzzy, I tried my hardest to get a clear one.)
(x+1)(x-1)(x2+4)
Now plug the x^2+4 into the quadratic formula.
In x2+4, a=1, b=0, and c=4.
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}={-0 \pm \sqrt{0^2-4(1)(4)} \over 2(1)}\)
Now simplify. -0 can just go away and so can 0^2.
\(x={ \pm \sqrt{-4(1)(4)} \over 2}\)
4*1*-4=-16
\(x={\pm \sqrt{-16} \over 2}\)
The square root of 16 is 4 and the square root of -1 is i.
\(x=\frac{\pm4i}{2}\)
This can be simplified to \(x=\pm2i\).
So, the factors are \((x+1)(x-1)(x+2i)(x-2i)\).
+895 1) For the first, the number of complex zeros you have is always the same as your highest degree.
So, for that one it would be 4 complex zeros.
2) f(x)=x4+3x2−4
Here is the graph: https://www.desmos.com/calculator/zm7f4iwvrb
From the graph, the two zeros are (-1,0) and (1,0).
Now, I'm going to use long division for these.
Pull off all the coefficients (Don't forget the missing terms. This problem is really x4+0x3+3x2+0x-4)
I'm gong to use (-1,0) first, since it makes x+1 as a factor.
(Sorry if the picture is a bit fuzzy, I tried my hardest to get a clear one.)
(x+1)(x-1)(x2+4)
Now plug the x^2+4 into the quadratic formula.
In x2+4, a=1, b=0, and c=4.
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}={-0 \pm \sqrt{0^2-4(1)(4)} \over 2(1)}\)
Now simplify. -0 can just go away and so can 0^2.
\(x={ \pm \sqrt{-4(1)(4)} \over 2}\)
4*1*-4=-16
\(x={\pm \sqrt{-16} \over 2}\)
The square root of 16 is 4 and the square root of -1 is i.
\(x=\frac{\pm4i}{2}\)
This can be simplified to \(x=\pm2i\).
So, the factors are \((x+1)(x-1)(x+2i)(x-2i)\).
