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# How many complex zeros does the polynomial function have?

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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)

Guest Nov 9, 2017

#1
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+1

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 $$(x+1)(x-1)(x+2i)(x-2i)$$.

#1
+771
+1

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 $$(x+1)(x-1)(x+2i)(x-2i)$$.