#1

How would I find the zeros for this function?

x^4 + x^3 - 6x^2 - 14x + 12

To find the zeros, would you basically factor it and use the quadratic equation if necessary to find the complex zeros? I tried factoring it using the factor theorem, but none of the factors of 12 work out.

x^4 + x^3 - 6x^2 - 14x + 12

To find the zeros, would you basically factor it and use the quadratic equation if necessary to find the complex zeros? I tried factoring it using the factor theorem, but none of the factors of 12 work out.

#2

polynomial division??????

#3

polynomial division??????

That's what I was just trying out. I'll post when I get results.

#4

polynomial division??????

?

What would I divide by? I've tried using the factor theorem, but none of the factors of 12 work out.

#5

Are you sure that it's +12 at the end? It would work out with -12.

#6

plug it into a graphing calculator and see where it crosses the x axis

#7

Crap, I learned this a couple months ago and I've already forgotten it.

#8

Hmm. I'll double check that. I'll ask a friend in a bit. If it was -12, x-3 would be a factor.

I need help with a similar problem. Find the other zeros, given that 3 + 2i is one of them.

x^4 - 8x^3 + 7x^2 - 38x + 26

3 - 2i is one for sure, but how would I find the others? Same process as the previous problem, as in working with factors of 26 until I find a real zero?

I need help with a similar problem. Find the other zeros, given that 3 + 2i is one of them.

x^4 - 8x^3 + 7x^2 - 38x + 26

3 - 2i is one for sure, but how would I find the others? Same process as the previous problem, as in working with factors of 26 until I find a real zero?

#9

#10

Divide it using 3+2i and 3-2i.

#11

Yeah, synthetic division is the way to go.

Also, fun fact: 2 zeros, not nice points.

Also, fun fact: 2 zeros, not nice points.

*Last edited by tylerishot at Apr 5, 2008,*

#12

Synthetic Division.....

We just did this in Algebra II a week ago so yeah..... I was almost like "Oh **** I forgot this!"

We just did this in Algebra II a week ago so yeah..... I was almost like "Oh **** I forgot this!"

#13

Hmm. I'll double check that. I'll ask a friend in a bit. If it was -12, x-3 would be a factor.

I need help with a similar problem. Find the other zeros, given that 3 + 2i is one of them.

x^4 - 8x^3 + 7x^2 - 38x + 26

3 - 2i is one for sure, but how would I find the others? Same process as the previous problem, as in working with factors of 26 until I find a real zero?

Done the second one. The roots are: 3+2i, 3-2i, 1+i, 1- i.

Working:

If 3+2i and 3-2i are given, then we can show the following using the polynomail theorem:

x^4 - 8x^3 + 7x^2 - 38x + 26 = (x-(3-2i))(x-(3+2i))(Ax^2 + Bx + C)

From here you expand out the first two brackets, you end up with:

(x^2 -3x -2ix -3x +9 +6i +2ix -6i +4)(Ax^2 + Bx + C)

Now look in the first bracket, and see what cancels:

(x^2 -6x +13)(Ax^2 +Bx +c)

Then expand this:

.... i'm not gonna write that out, it takes ages.

Then equate the coefficients:

x^4 - 8x^3 + 7x^2 - 38x + 26 = Ax^4 +(B-6)x^3 + (C-6B+13)x^2 + (13B-6C)x +13C

From the first coefficient you can see that A=1 so that's why i didn't include A in the other coefficients.

So

B-6 = -8

13C = 26 So C = 2 and B = -2

And this is where i think you made a mistake in writing out the problems equation: 7x^2 should be 27x^2 methinks. Otherwise it's bullsh*t.

Anyway, then you go back a few steps and substitute A,B and C:

x^4 - 8x^3 + 7x^2 - 38x + 26 = (x-(3-2i))(x-(3+2i))(x^2 -2x + 2)

Then use the quadratic formula on the second quadratic, resulting in two more roots:

1+i and 1- i.

Yea?

#14

I'm still learning quadratic equations >_>

#15

btw for the first one, it's too hard to do straight off the bat like that. I did A level Cambridge and the hardet question like that that they could ask you would have at least one 'nice' root like 1,-1,0,2, or -2. They would then expect you to use the polynomial theorem.

#16

i just had to do an assignment on this. i think you set x to zero, and find the dilation factor...

im dropping to maths a next term cuz i dont really need maths b, and cuz im sh*t at it... hahaha

im dropping to maths a next term cuz i dont really need maths b, and cuz im sh*t at it... hahaha

#17

The equation is to the Fourth power so it has 4 zeros. Whenever an i is involved the oppisite of i is also. Use synthetic division to devide both of those out. When you are finished use the quadratic formula.

#18

graphing calculator? you can probably find some kind of freeware calculator on the net even.

anyways,

x= .691 , 2.63

anyways,

x= .691 , 2.63

#19

graphing calculator? you can probably find some kind of freeware calculator on the net even.

anyways,

x= .691 , 2.63

ARe those exact answers or estimations.

#20

those are exact answers unless you need them to like what? 10 decimal places?

#21

those are exact answers unless you need them to like what? 10 decimal places?

well he should prob use the exact answer (fractions). He could use the decimals and see if they are the same value of any of the fractions you get by taking p over q

#22

graphing calculator? you can probably find some kind of freeware calculator on the net even.

anyways,

x= .691 , 2.63

those are exact answers unless you need them to like what? 10 decimal places?

nah man they're quite off aye. If you sub 2.63 in, you get -0.28644739 which is way off.

Besides, we want algebraically elegant roots and methinks at least two are complex.

#23

Okay! I screwed up the problems big time! lol Sorry!

1) x^4 + x^3 - 6x^2 - 14x

I can do that one on my own.

2) x^4 - 8x^3 + 27x^2 - 38x + 26

3+2i is one zero.

Hehe, thanks again to anyone who helped, but I still don't quite get the process to solving number 2. Could someone please explain it a bit differently?

1) x^4 + x^3 - 6x^2 - 14x

**-**12I can do that one on my own.

2) x^4 - 8x^3 + 27x^2 - 38x + 26

3+2i is one zero.

Hehe, thanks again to anyone who helped, but I still don't quite get the process to solving number 2. Could someone please explain it a bit differently?

#24

Can you believe our curriculum got rid of complex numbers?

I'm in grade 12 and I haven't got a clue.

I'm in grade 12 and I haven't got a clue.

#25

Okay! I screwed up the problems big time! lol Sorry!

2) x^4 - 8x^3 + 27x^2 - 38x + 26

3+2i is one zero.

For number 2, use synthetic division with (3+2i) and then once more with (3-2i). It'll get messy at points, but it'll neaten up towards the end. If you don't know how to do synthetic division, google it. I could explain in person, but we don't have that luxury at the moment.

Here's something to get you started.

http://www.purplemath.com/modules/synthdiv.htm

EDIT: Grammar

#26

For number 2, use synthetic division with (3+2i) and then once more with (3-2i). It'll get messy at points, but it'll neaten up towards the end. If you don't know how to do synthetic division, google it. I could explain in person, but we don't have that luxury at the moment.

Here's something to get you started.

http://www.purplemath.com/modules/synthdiv.htm

EDIT: Grammar

Ah I got it. Thanks for the suggestion, but I found a neater way of doing it. Multiply the two solutions together to get the coefficients for the quadratic in the form of c/a.

(3+2i)(3-2i) = 9-4i^2 =13/1.

So c=13 and a =1

and

b/a = -([3+2i]+[3-2i]) = -6/1 So b=-6

Put that together to get (x^2 - 6x + 13)(some quadratic) = x^4 - 8x^3 + 27x^2 - 38x + 26. Then piece together the quadratic to get...

(x^2 - 6x + 13)(x^2 -2x + 2)= x^4 - 8x^3 + 27x^2 - 38x + 26

Use the quadratic formula with x^2 - 2x + 2, and you get

**2±i**.

#27

Ah I got it. Thanks for the suggestion, but I found a neater way of doing it. Multiply the two solutions together to get the coefficients for the quadratic in the form of c/a.

(3+2i)(3-2i) = 9-4i^2 =13/1.

So c=13 and a =1

and

b/a = -([3+2i]+[3-2i]) = -6/1 So b=-6

Put that together to get (x^2 - 6x + 13)(some quadratic) = x^4 - 8x^3 + 27x^2 - 38x + 26. Then piece together the quadratic to get...

(x^2 - 6x + 13)(x^2 -2x + 2)= x^4 - 8x^3 + 27x^2 - 38x + 26

Use the quadratic formula with x^2 - 2x + 2, and you get2±i.

That is a very odd way to solve that. Never in my 4 years of algebra have I seen it done that way. :-\ pretty cool though i guess

#28

That is a very odd way to solve that. Never in my 4 years of algebra have I seen it done that way. :-\ pretty cool though i guess

Hehe. I'm just learning this, so I wouldn't know if that's weird or not. I don't know how to do synthetic division with imaginary numbers, so yeah

#29

Ah I got it. Thanks for the suggestion, but I found a neater way of doing it. Multiply the two solutions together to get the coefficients for the quadratic in the form of c/a.

(3+2i)(3-2i) = 9-4i^2 =13/1.

So c=13 and a =1

and

b/a = -([3+2i]+[3-2i]) = -6/1 So b=-6

Put that together to get (x^2 - 6x + 13)(some quadratic) = x^4 - 8x^3 + 27x^2 - 38x + 26. Then piece together the quadratic to get...

(x^2 - 6x + 13)(x^2 -2x + 2)= x^4 - 8x^3 + 27x^2 - 38x + 26

Use the quadratic formula with x^2 - 2x + 2, and you get2±i.

Wrong. Using the quadratic formula you get 1+/- i. But yea, if you take a look at the solution i provided yesterday on page 1, you would see that not only did i solve it for you, but i even told you that you misprinted the question wrong

To remind you of what i wrote:

Done the second one. The roots are: 3+2i, 3-2i, 1+i, 1- i.

Working:

If 3+2i and 3-2i are given, then we can show the following using the polynomail theorem:

x^4 - 8x^3 + 7x^2 - 38x + 26 = (x-(3-2i))(x-(3+2i))(Ax^2 + Bx + C)

From here you expand out the first two brackets, you end up with:

(x^2 -3x -2ix -3x +9 +6i +2ix -6i +4)(Ax^2 + Bx + C)

Now look in the first bracket, and see what cancels:

(x^2 -6x +13)(Ax^2 +Bx +c)

Then expand this:

.... i'm not gonna write that out, it takes ages.

Then equate the coefficients:

x^4 - 8x^3 + 7x^2 - 38x + 26 = Ax^4 +(B-6)x^3 + (C-6B+13)x^2 + (13B-6C)x +13C

From the first coefficient you can see that A=1 so that's why i didn't include A in the other coefficients.

So

B-6 = -8

13C = 26 So C = 2 and B = -2

And this is where i think you made a mistake in writing out the problems equation: 7x^2 should be 27x^2 methinks. Otherwise it's bullsh*t.

Anyway, then you go back a few steps and substitute A,B and C:

x^4 - 8x^3 + 7x^2 - 38x + 26 = (x-(3-2i))(x-(3+2i))(x^2 -2x + 2)

Then use the quadratic formula on the second quadratic, resulting in two more roots:

1+i and 1- i.

Yea?

#30

Wrong. Using the quadratic formula you get 1+/- i. But yea, if you take a look at the solution i provided yesterday on page 1, you would see that not only did i solve it for you, but i even told you that you misprinted the question wrong

To remind you of what i wrote:

Yep, I saw it I didn't get why you were using linear factors though, but we went over it today. Thanks for pointing out my error with dividing 2 by 2 Thanks for the help!