Need math help! Precalculus

[jaydee]

These are model problems for part of tomarrows final exam.

Problem #1:
Find a polynomial function whose zeros are +3, +2, -1.

Problem #2: Find all zeroes of the function f(x) = 2x^3 - 7x^2 - 17x + 10

Problem #3:
Find two fractions whose sum/difference is:
(3x-11)/(x^2+x-6)

Thanks a lot guys, I appreciate it.

 

[GoldenBear]

Problem #1:
Find a polynomial function whose zeros are +3, +2, -1.

How about (x-3)(x-2)(x+1) expanded..Is a polynomial function one with x^2? If so never mind..

Problem #2: Find all zeroes of the function f(x) = 2x^3 - 7x^2 - 17x + 10

I think that's a problem that involves p's and q's and all..and I don't remember any of that

Problem #3:
Find two fractions whose sum/difference is:
(3x-11)/(x^2+x-6)

That and zero?

 

[jaydee]

So GoldenBear, would that make it x^3-4x^2+x+6?
Oh, and I don't think thats quite was he's looking for on #3 heh.

 

[GoldenBear]

Yes that's what I get..

 

[jaydee]

Cool. One down, two to go...

 

[marketsons1985]

For Problem #2.....
If you are allowed to use a graphing calculator,

Plug it in and solve!

 

[DAM]

for number two, just take the derivative and make it equal to zero.






dam()

 

[jaydee]

The answer only gives me 1 point out of six, I gotta show work long hand for full credit. Yes I do have a TI-86.

 

[RossGr]

Number #3
3x - 11/(x^2 + x -6) = 3x/(x^2 + x -6) - 11/(x^2+x-6)


since x^2+x-6 = (x-2)(x+3) pick any value of x <> 2 or -3
and and plug in to the the sum.

YOU will need to use either synthetic division or a graphing calculator to find the roots of #2.

 

[marketsons1985]

Dam....
We just finished derivatives, and unless I'm wrong, when you do that, it only gives you the relative minimums and maximums of the origninal equation....

 

[marketsons1985]

Oh yeah...I just remembered, if you've got the 1 point from the calculator, then just do what Ross said and synthetically divide it.

 

[jaydee]

Thats all? I can do that. Synthetic division is actually something I remember how to do. Thanks

 

[jaydee]

Allright, so how do I find the derivative on #2 then?

 

[Mday]

#2, any real roots of the function would be divisors of 10. so, for simplicity, try integers. or you can use the cubic formula.

knowing the derivative does not quite help you.

--

this is precalc? wtf man.

 

[jaydee]

Thanks a lot Mday. I got it figured using trial and error synthetic division to get 5 as one root, and factored the remaining 2x^2+3x-2 to be (2x-1)(x+2), so I have 5, 1/2, and -2 for answers.

RossGr, after further investigation, I don't understand what you do with either (x+3)(x-2) to get to synthetic division in order to get the sums/differences. Is this one where I use the partial fractions theorum?

 

[jaydee]

Ok, I used the partial fractions method and this is what I got:

4/(x+3) + (-1)/(x-2)

Someone please confirm this!!!!!

 

[Mday]

jaydee, #3 is a precursor to what is called partial fractions in integral calculus.

as you can see, you have a function which can be represented as f(x) / [(g(x) * h(x)] where f = 3x-11, g = x+3, h = x-2

so, if you recall adding fractions with numbers as the num/den , the final result would have some combination of the denominators of the fraction. for example: 1/2 + 1/3 = 3/6 + 2/6 = 5/6.

notice how g(x) and h(x) are what is called irreducible, that is, they CANNOT be seperated as they are the most primative NON integer polynomials (monomials).

now let's do this backwards: x/(x+1) + x/(x+2) = ??

you would do this by x(x+2)/(x+1)(x+2) etc... right?

well, as you can see, this is similar to what you do with #3.

setup this:

A/x+3 + B/x-2

where A and B are some rational numbers. when doing such problems, they wind up being fractions.

note, if the denominator of one of factors happen to be of higher degree, say the denominator is (xx+1)(x-1), you have to do this: (Ax+B)/(xx+1) + C/(x-1) as to determine what the actual functions are.

 

[Mday]

oh, work out:

A/(x+3) + B/(x-2) and set that to what you want.

solve for A and B, and you have an answer =)

 

[Mday]

typing = slow...

anyway, that is the correct answer.

 

[jaydee]

Mday, yeah thats what I did. I ended up with B=-1, and A=4 after going through all that and then substitution to solve for B, then A. It checkso out right, so I'm assuming, 4/(x+3) + -1/(x-2) are right. Thanks a lot I appreciate the help greatly.

 

[jaydee]

All Your Partial Fraction Decompostion Are Belong To Us.... Bwahahahahahaa

 

[nd]

You have a TI-86? Here's how I solved the first two using a TI-86:

Find a polynomial function whose zeros are +3, +2, -1.

Of course, many functions will satisfy these requirements. You realize it will follow the form y = Ax^3 + Bx^2 + Cx + D. Choose another test point, like (0,2). Plug these into a table and do a power 3 regression on them. Do this by hitting [2nd], then STAT. F2 (edit), and put in 3, 2, and -1 for the xStat column, and 0's for the yStat column. Finally, add in your test point of x=0 and y=2. Hit Exit. Now you can perform the power regression by hitting [2nd] STAT again, then choose F1 (calc), hit MORE, then hit F5 for P3Reg. Hit exit twice, then [2nd] LIST, F3 (names), and select xStat, type a comma, and then yStat (basically, you want P3Reg xStat,yStat to appear). Hit enter and it will give you an equation to satisfy the table. Scroll right to see the whole answer. In this case, you will get y=(1/3)x^3 - (4/3)x^2 + (1/3)x + 2.

Problem #2: Find all zeroes of the function f(x) = 2x^3 - 7x^2 - 17x + 10

Hit [2nd] POLY. Choose 3 for order. then a3=2, a2=-7, a1=-17, a0=10, then hit F5 to solve. You will see that the answer is x = {5, -2, (1/2)}

I went through synthetic division and all that other stuff in pre-calc too. Once you get in higher maths it becomes tedious and you rely on your calculator more

 

[Mday]

<< Once you get in higher maths it becomes tedious and you rely on your calculator more >>



IF THEY LET YOU...

i can't factor much for my life. and imagine that during my algebra class, we were doing galois theory, and i can't factor =(

hehe, almost called a polynomial irreducible when it was not, threw away a whole page of a proof and continued... heheheee...

 

[Mday]

actually, come to think of it, i did do this stuff in precalc. that was 5 years ago

of course i was taking calculus at the time... guess which i am better at? =)

 

[RossGr]

nd you are making a simple problem way hard.

When you are given the roots and want to create a Polynomial

say roots at a, b, and c. You know that the factored form will look like (x-a)(x-b)(x-c). If you want the sum of powers form multiply the factors. This example (3 roots) will result in a cubic.

I guess I understood the earlier problem I laid out. I factored the denominator as you do not want it to be zero. So if you avoid its roots you should be able plug in integers and get different fractions whos sum/difference is the original expression. To a mathmatician sum and difference is the same operation so I am a bit confused by the differenation expressed here.