Finding degree-3 polnomial from roots: 1 + 4i, 1 - 4i, -1
- Thread starter amlee
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
Re: Finding polnomials
You should know this: (a+b)(a-b) = a^2 - b^2
This is very important for multiplying Complex Numbers.
Use your algebra and multiply the first two factors galactus gave you.
(x-(1+4i))(x-(1-4i))
What do you get?
You should know this: (a+b)(a-b) = a^2 - b^2
This is very important for multiplying Complex Numbers.
Use your algebra and multiply the first two factors galactus gave you.
(x-(1+4i))(x-(1-4i))
What do you get?
Re: Finding polnomials
If you need help UNDERSTANDING, then make use of facts your teacher has presented, such as:
If you have a quadratic with real coefficients, complex roots occur as conjugate pairs.
In a quadratic ax^2 + bx + c = 0, the sum of the roots is -b/a and the product is c/a
To multiply COMPLEX conjugates, a +- bi, the product is just a^2 + b^2.
........................
Now put a couple of things together. (Yes, yes, you are producing a cubic, but you don't have to do it all at once.)
Get the quadratic factor:
1+4i, 1-4i are the roots.
Their sum is 2. If we assume a = 1, and we can, then b = -2.
Their product is 17. Then c = 17.
So the quadratic factor is x^2 - 2x + 17.
Now she also taught you that if x = r is a root, then (x - r) is a factor.
So x + 1 is a factor.
Your polynomial is the product of those two. And you don't have a ton of algebra to do when you use these facts.
amlee said:
If you need help UNDERSTANDING, then make use of facts your teacher has presented, such as:
If you have a quadratic with real coefficients, complex roots occur as conjugate pairs.
In a quadratic ax^2 + bx + c = 0, the sum of the roots is -b/a and the product is c/a
To multiply COMPLEX conjugates, a +- bi, the product is just a^2 + b^2.
........................
Now put a couple of things together. (Yes, yes, you are producing a cubic, but you don't have to do it all at once.)
Get the quadratic factor:
1+4i, 1-4i are the roots.
Their sum is 2. If we assume a = 1, and we can, then b = -2.
Their product is 17. Then c = 17.
So the quadratic factor is x^2 - 2x + 17.
Now she also taught you that if x = r is a root, then (x - r) is a factor.
So x + 1 is a factor.
Your polynomial is the product of those two. And you don't have a ton of algebra to do when you use these facts.
- Joined
- Feb 4, 2004
- Messages
- 16,582
If the classroom lectures and your textbook have left you still unable even to being to follow the instructions here, then one more worked example obviously isn't going to "do the trick". You need lessons.amlee said:
While we cannot provide the course lectures requested, there are loads of great lessons available online:
. . . . .Google results for "polynomial from zeroes work backwards"
Please study at least two lessons from the above link. Once you have learned the basic terms, concepts, and techniques, please attempt the exercise. If you get stuck, you will then be able to reply with a clear listing of your work and reasoning so far.
Thank you!
Eliz.