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I have been assigned a type of math problem that I have almost no idea how to complete. I think I missed something basic a long time ago... I was hoping that someone could show me how it's figured out so that I can solve this stuff.

example:

textbook wrote:Find polynomial with real coefficients that meets these conditions:

Degree 4; zeros at x = 3, x = -1, and x = 2; f(0) = 30

Any help would be greatly appreciated.

f(x)=c4x^4+c3x^3+c2x^2+c1x +c0

f(0)=30=>c0=-30

brainfart deleted.

replace x by every known zero and you have a three equations, three variables system. Simply solve it.

erm, the highest exponent will be 4, and the last term will be positive 30. I don't know how to do the rest (damn you, grade 11 math!)

Step 1:
With the zeroes, you know it takes the form of something times (x-3)(x+1)(x-2), which equals zero at x = 3, -1, and 2.
2:
Then set x to 0 and multiply those out to get something times six equals 30.
3:
Five times six equals 30, so your polynomial is (x+5)(x-3)(x+1)(x-2). Just multiply all that that out and you're done.

(x-3)(x+1)(x-2) ....here are your roots, so thats given

assuming you can't have extra roots somewhere:

Well, I don't think this gives you an if and only if polynomial but since we already have 3 roots and we don't want any weird imaginary roots or something, it is easiest to make a double root or something.

(x-3)(x+1)(x-2)(x+1)a

now to find varible "a" using boundary conditions

(0-3)(0+1)(0-2)(0+1)a = 30

a = 5

One polynomial that fits is the product of (x-3)(x+1)(x+1)(x-2)



assuming you can have extra roots somewhere: (it probably isn't the question, but I typed this first and didn't want to delete it)

degree four so we can mutiply another x somewhere. Since we have a boundary condition, we'll plug it in. While there isn't enough boundary conditions to give an if and only if polynomial, the question is probably only asking for one that fits so we'd just plug in another (ax + b) term. Since x is at zero, a can be any value where well assign one.

(0-3)(0+1)(0-2)(0 + b) = 30

b = 5

Colonel Olrik wrote:f(x)=c4x^4+c3x^3+c2x^2+c1x +c0

f(0)=30=>c0=-30

brainfart deleted.

replace x by every known zero and you have a three equations, three variables system. Simply solve it.

Acutally, while the approach is probably the most "valid", you have 5 unknowns, giving alot of leeway on what crazy numbers you want to plug in.

So in the end you have 5 unknowns and 4 equations and infinitely many solutions!

Gah linear algerbra

SWPIGWANG wrote: So in the end you have 5 unknowns and 4 equations and infinitely many solutions

ugh, I misread it once and still didn't get it right the second time. The easiest solution is the one you and others have presented. I'm going to bed now.

[sigh, there was a time I could solve this shit in two seconds with one week of sleep deprivation and very, very drunk]

f(x)=5x^3-20x^2+5x+30




It's been a while. Tell me if I'm wrong.

Wicked Pilot wrote:f(x)=5x^3-20x^2+5x+30




It's been a while. Tell me if I'm wrong.

OK, I think I'm sober now.

You're wrong because it must be a fourth order equation. So, you have:

f(x)=(x-3)(x+1)(x+1)(x-5)*5

Now it's of fourth order and obeys all conditions, having a double zero in -1 (one possible option).

f(x)=c4x^4+c3x^3+c2x^2+c1x +c0

unknowns, c0, c1, c2, c3, c4 which is 5 unknowns

f(0)=30
f(3)=0
f(2)=0
f(-1)=0

4 equations.

Expanded out

f(0) = 30 = c0
f(3) = 0 = 81(c4) + 27(c3) + 9(c2) + 3(c1) + c0
f(2) = 0 = 16(c4) + 8(c3) + 4(c2) + 2(c1) + c0
f(-1) = 0 = 1(c4) - 1(c3) + 1(c2) - 1(c1) + c0

Solving gives you the solution space to all the answers muhahahahahahahahahahahahahahahahahahahahahahahah

muhahahahahahahahahahahahahahahahahahahahahahahahaha

MUHAHAHAHAHAHAHAHAHAHAHAHAHAHA

I hate math

SWPIGWANG wrote:f(x)=c4x^4+c3x^3+c2x^2+c1x +c0

unknowns, c0, c1, c2, c3, c4 which is 5 unknowns

f(0)=30
f(3)=0
f(2)=0
f(-1)=0

4 equations.

I know that, you sod, I made an elementary math mistake once in a lifetime. Enjoy it, you won't catch me making another for a long time

Actually (and I really must go to bed) my original solution is the best, since the equations do solve the problem, by giving all solutions. Any possible particular solution can be obtained by assigning one number to a variable and solving the three equations for the three remaining variables.

Thank you everyone for your help! I am fairly confident on how that works now, but I am having difficulty applying it to these problems that involve imaginary numbers (I know nothing about i except that I used it in a java fractal-producing program once a long time ago w/o knowing its true purpose)...

same question as before but with these conditions wrote:Degree 4; zeros at x = 1 - 2i and x = 1 + i; f(0) = 20

If I can figure out how to do this final type, and then I'm set for the year on math. Sorry to continue asking these annoying questions! It's only that people here are way better at explaining things than the teacher...

Seggybop wrote:Thank you everyone for your help! I am fairly confident on how that works now, but I am having difficulty applying it to these problems that involve imaginary numbers (I know nothing about i except that I used it in a java fractal-producing program once a long time ago w/o knowing its true purpose)...

same question as before but with these conditions wrote:Degree 4; zeros at x = 1 - 2i and x = 1 + i; f(0) = 20

If I can figure out how to do this final type, and then I'm set for the year on math. Sorry to continue asking these annoying questions! It's only that people here are way better at explaining things than the teacher...

It's exactly the same trick, with one twist. You must know that complex roots always come in conjugate pairs, so if you have a zero in 1+i then you will also have one in 1-i. Those 4 zeros define the fourth degree equation.

So, the solution to the problem is obviously

f(x) = (x^2+4)(x^2+1)*5

Since by definition i = square root (-1)

I want my hour of sleep back

Seggybop wrote:I have been assigned a type of math problem that I have almost no idea how to complete. I think I missed something basic a long time ago... I was hoping that someone could show me how it's figured out so that I can solve this stuff.

example:

textbook wrote:Find polynomial with real coefficients that meets these conditions:

Degree 4; zeros at x = 3, x = -1, and x = 2; f(0) = 30

Any help would be greatly appreciated.

Ok, hold on. I will try not to do the work for you, but show you how to do it. (I'll put the answers in "spoiler" format.)

Find polynomial with real coefficients that meets these conditions:

Degree 4; zeros at x = 3, x = -1, and x = 2; f(0) = 30

So, for each zero, you know that you have a factor.
I.e., x=3 => (x-3) is a factor.

So, here you have three factors from the three zeroes, plus another one which you don't know, but must be a duplicate of one of the other factors (since degree is 4 and you only have 3 zeroes). So you want something of the form:

ax^4 + bx^3 + cx^2 + dx^1 + e = f(x)

Then you want to set it up so that f(0)=30 That means that everything "times" an x is now "times" a 0.
so f(0)= a(0) +...+ d(0) +e = e = 30.

Now you need to multiply your factors.
You get x^3 - 4x^2 + x + 6, unless I made a mistake.

To keep your e coefficient positive you need to multiply by a positive factor (like (x + 1)) to get a fourth degree polynomial.
--Are you sure you didn't leave something out?

This gives us
x4 - 3x3 - 3x2 + 7x + 6

Then you will need to multiply the entire equation by an appropriate number (like 5) to get the 0th power coefficient to be 30.

Then you have:
5x4 - 15x3 - 15x2 + 35x + 6[/size]

Seggybop wrote:Thank you everyone for your help! I am fairly confident on how that works now, but I am having difficulty applying it to these problems that involve imaginary numbers (I know nothing about i except that I used it in a java fractal-producing program once a long time ago w/o knowing its true purpose)...

same question as before but with these conditions wrote:Degree 4; zeros at x = 1 - 2i and x = 1 + i; f(0) = 20

If I can figure out how to do this final type, and then I'm set for the year on math. Sorry to continue asking these annoying questions! It's only that people here are way better at explaining things than the teacher...

You can use the same method as I told you about in my previous post, but just remember that i^2 = -1

and to cancel out a (1 + i) you must multiply by (1 - i) and (1-2i) with (1 + 2i) etc.
The equation you get will multiply each zero with its conjugate pair, i.e:
(x - 1 + 2i)(x - 1 -2i)

Please let me know if my comments are helpful or confusing.

I'm currently a professional tutor and I'm getting my credential to be a teacher, so please let me know if I'm doing a good job.