Question

find all roots of the polynomial
p(z) = 2z^3-2z^2-z-6

Answer 1

cubic equation omg

Answer 2

z = x+iy this is complex number question

Answer 3

Well first of all you can find out that P(2)=0

Answer 4

Therefore z-2 must be a factor of 2z^3 - 2z^2 - z - 6

Answer 5

There will be one + ve real root.)

Answer 6

why is it that you have fed 2 into the equation, is this a random number

Answer 7

When you try to factorize polynomials you can try plugging in numbers to see which one of them give you a zero,
it is a trial and error way to factorize easily

Answer 8

So if P(2) gives you zero, z-2 must be a factor of p(z)

Answer 9

Then when you divide 2z^3 + 2z^2-z-6 by z-2

Answer 10

You get 2z^2 + 2z +3

Answer 11

Sum of the Roots , product of the roots

Answer 12

So we have the following expression (2z^2 +2z +3)(z-2)=0

Answer 13

Now we have found one solutions z = 2

Answer 14

Other two are given by 2z^2+2z+3=0

Answer 15

and the other sol^n will be Complex numbers...

Answer 16

Gud work...@baddinlol

Answer 17

I am going to complete the square

Answer 18

polynomials are fun aren't they

Answer 19

nop...@baddinlol use quadratic formula

Answer 20

2(z^2 + z + 3/2) = 0
z^2 + z + 3/2 = 0
(z + 1/2)^2 + 5/4 = 0

Answer 21

(z + 1/2)^2 = -5/4
z + 1/2 = sqrt(5)i /2, -sqrt(5)i
/2
z = (sqrt(5)i - 1)/2, (-1-sqrt(5)i)/2

Answer 22

that will be =+or-sqrt b^2-4(a)(c)/(2a)

Answer 23

well done baddinol thanks

Answer 24

np!

Answer 25

Did you know about putting in numbers to find p(Z) = 0?

Answer 26

By the way the numbers you plug in have to be a factor of the constant in the equation, which in your equation is -6::::: 2z^3-2z^2-z-6

Answer 27

i did but i thought there must be a faster way

Answer 28

SO i tried 1, -1, 2 and when i got upto 2 i stopped because i found P(2)=0

Answer 29

Are you allowed to use a cas calculator?

Answer 30

no