| Author | Comment |
Vectris
Ultralisk
Posted: 10 Oct 2008
15:03 GMT
Total Posts: 375 | I doubt this will even be graded for accuracy, and I'll ask my teacher on Monday anyways, but I got stuck on this problem and was wondering if anyone could help.
The problem is finding a polynomial function when you are given the number of solutions for it.
The solutions given are: -2, -2, and -5i
So naturaully I put them = to x and transfer them over to the x side so I can multiply them and get the polynomial.
(x + 2)(x + 2)(x + 5i)
Well first I multiplied the x + 2s to get:
(x^2 + 4x + 4)(x + 5i)
I was about to do the next step, multiplying them, when I realized that the 5i isn't going to cancel out when I multiply it, and we're not supposed to have 5ix or anything. When we did these in class there was always a negative in the multiplund or at least another i and they would square making a negative. But no, in this problem, if I multiply them, I'm going to get some ix numbers no?
I looked in the back of the book and the answer contains no ix numbers.
x^4 + 4x^3 + 29x^2 + 100x + 100
So what happened to the xis??? The answer says there are none but when I start to multiply I immediately get one.
(x^3 + 4x^2 + 4x)(x + 5i)
I multiply 5i by x^3 and get... 5ix^3, wallah, a xi. Well the only way I can think of that they didn't keep the xis are that the ^3 from the x goes on the i too making the i a positive one. But I swear that it doesn't work like that, the exponent from the x doesn't just jump to the i.
Anyone have any ideas? |
Zachary940
Wraith
Posted: 10 Oct 2008
17:32 GMT
Total Posts: 714 | 1st off (x+2)(x+2) is (x^2+4x+4)
So I say type o in the book, because there is no way to get a x^4.
Yes I have seen type o's in math books.
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It is much easier to suggest solutions when you know nothing about the problem. |
Vectris
Ultralisk
Posted: 10 Oct 2008
17:52 GMT
Total Posts: 375 | Oops, ya, I had it right on my paper but I typed it out with an extra degree to x.
But ya, your still right. It comes close as the -5i goes to x = 5i, so there is a x^3, but ya there's no x^4.
I suppose it is a typo. I double checked the question and made sure I was looking at the right answer in the book.
Just in case I misinterpreted the question here it is:
Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.)
And of course the given zeros where:
-2, -2, -5i
One other thing is that since there's two -2s, we know -2 is a double root, and I can't remember if you do anything special with double roots, still I think setting them all with x and then multiplying should still get you the right answer. |
Zachary940
Wraith
Posted: 11 Oct 2008
11:53 GMT
Total Posts: 714 | you forgot 5i, imaginary numbers always come in pairs.
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It is much easier to suggest solutions when you know nothing about the problem. |
Vectris
Ultralisk
Posted: 11 Oct 2008
13:51 GMT
Total Posts: 375 | Ah yes that's right. Although isn't it that whenever you have x^2 = something that by removing the ^2 (squaring both sides) that's what gives you the +-. And since the only way to get i is by the square root of a negative number, that means x^2 was equal to some negative number. Anyways, you get the two is from squaring both sides, +i and -i.
Thanks for figuring it out! That explains the x^4 and how to cross out the "i"s. I'm about to try and solve it now, I'm sure I'll get it, and I don't remember the answer so I can hopefully still gain practice from this.
EDIT: I was just wondering, why wouldn't the book give you both of the "i"s though?
Also, there is one more problem on just like this and it has one of the zeros as 1 + the square root of 3 i. So that means there's a 1 - the square root of 3 i, right? |
Zachary940
Wraith
Posted: 12 Oct 2008
10:47 GMT
Total Posts: 714 | It is one of those "you should know" things.
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It is much easier to suggest solutions when you know nothing about the problem. |
