Anonymous

# Math problem help - finding real and imaginary zeros of a polynomial?

My math teacher assigned us some homework over the winter break and I've had some trouble with three of the questions. We are supposed to find all real and imaginary zeros to the given polynomials:

24x^5 + 3x^2 factors to --> 3x^2(2x+1)(4x^2 - 2x +1) (?)

16x^4 - 1 factors to --> (2x+1)(2x-1)(4x^2+1)

2x^4 + 5x^3 - 3x^2 - 4x + 2

Now I've found out the real zeros alright but I can't figure out the imaginary ones. Can anyone help me?

Relevance

Hi,

24x^5 + 3x^2 factors to --> 3x^2(2x+1)(4x^2 - 2x +1) (?)

3x² (2x+1)(4x² - 2x +1) = 0

3x² = 0

2x + 1 = 0

4x² - 2x +1 = 0

. . . . . .___________

-(-2) ± √(-2)² - 4(4)(1)

------------------------------ =

. . . . . 2(4)

. . . ._____

2 ± √4 - 16

----------------- =

. . . . .8

. . . ._

1 ± i√3

. . .4

16x^4 - 1 factors to --> (2x+1)(2x-1)(4x^2+1)

(2x+1)(2x-1)(4x² +1) = 0

2x+1 = 0

x = -½

2x-1 = 0

4x² +1 = 0

4x² = -1

x² = -1/4i

. . . . . ____

x = ± √-1/4

2x^4 + 5x^3 - 3x^2 - 4x + 2 = 0

. ._____________

-1)2 . 5 . -3 . -4. .2

. .___-2_-3.__6_-2_

. . 2 . 3. -6 . . 2 . 0 <=shows -1 is a root

. ._____________

½)2 . 3. -6 . . 2

. .___1__2__-2__

. . 2 . 4 .-4. . .0 <=shows ½ is a root

2x² + 4x - 4 = 0

2(x² + 2x - 2) = 0

. . . . . ___________

-(2) ± √(2)² - 4(1)(-2)

------------------------------ =

. . . . . 2(1)

. . . . _____

-2 ± √4 + 8

------------------ =

. . . . 2

. . . . . _

-2 ± 2√3 . . . . . .__

------------ = -1 ± √3 <=ANSWER

. . . . 2

I hope that helps!! :-)

• Finding Imaginary Zeros

• The quadratic formula (which we all know and love) will predict the roots of a quadratic.

The cubic and quartic formulae are known in theory, but I doubt anyone here has them memorized or even has them available. It's too easy to ask a computer for the roots and have it rattle them off these days.

There's no quintic or higher degree formula which will solve a general polynomial of degree 5 or higher and there never will be. It's been shown that there are roots of 5th order polynomials that can't even be written in terms of roots! If you want to read up on this, the area that it's covered in is called Galois Theory. It's usually senior college or first year graduate work in mathematics to understand it.

• for the first one, just take the second quadratic (4x^2 - 2x +1) and plug it into the quadratic formula

(4x^2 - 2x +1)

x= 2+ or - sqrt(4-4*4) / 8

x=2+ or - sqrt(-12) / 8

x=2 + or - 2i sqrt(3)/8

so x= 1 + or - i sqrt(3) / 4

for the second one, take (4x^2+1)

set it equal to 0

4x^2+1=0

move the 1 over and divide by 4

x^2=-1/2

take the sqare root of both sides

you should get

x= + or - i/2

so your roots for that part are 0 + i/2 and 0 - i/2

• You are correct. There are no real zeros. Notice that coefficients for all the terms in the polynomial are positive. Notice also that all the powers of x are even. So none of the terms can be negative. And since the constant is positive, the expression will always be positive for all real x. So there are no real zeros.

• 24x^5 + 3x^2 factors to --> 3x^2(2x+1)(4x^2 - 2x +1)

real roots are 0, -1/2

imaginary is found this way

x= [2+/- sqrt (4-16)]/8

x= [2+/- sqrt (-12)]/8

x= [2+/- 2i sqrt (3)]/8

x= [1+/- i sqrt (3)]/4

------------------------------------------------------------------------

16x^4 - 1 factors to --> (2x+1)(2x-1)(4x^2+1)

so the real roots are 1/2 and -1/2 and the imaginary is found this way

4x^2 + 1 =0

4x^2 =-1

x^2 =-1/4

x= +/- i/2

------------------------------------------------------------------------------

• Anonymous

You're not going to have any imaginary numbers unless you have a negative square root so I don't see why he's asking you that..

I could be (and probably am) wrong though..

:) tee hee

oh yeah and -1 would equal i squared if that helps at all..

good luck with it. =D

• last factors (x+1)(2x-1) ( x² +2x-2)...again use the quadratic formula....you do know the Rational Root Theorem????