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anonymous
 3 years ago
Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = x^4 + 3x^3  5x^2  9x  2
anonymous
 3 years ago
Find a rational zero of the polynomial function and use it to find all the zeros of the function. f(x) = x^4 + 3x^3  5x^2  9x  2

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campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.1try f(1) ...so if f(1) = 0 then x = 1 is a rational zero... this uses the rational zero theory..... p/q p = factors of the constant q = factors of the leading term

campbell_st
 3 years ago
Best ResponseYou've already chosen the best response.1to find all the zeros... use synthetic division or polynomial division dw:1360356521817:dw

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1@gjhfdfg Do you know the rational root theorem? If you have some polynomial \[P(x) = a_nx^n+a_{n1}x^n1+\dots+a_1x^1+a_0\] and all the coefficients are integers, and \(p/q\) is a rational zero (and \(p/q\) is reduced), then \(p\) is a factor of \(a_0\) and \(q\) is a positive factor of \(a_0\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have done the rational root theorem but I never got it.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Also, if \(a_n =1\) then all rational zeros are integers which divide \(a_0\).

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1So here \(a_0 = 2\) and that leaves only 4 roots to try: \(1/1, 2/1, 1/1, 2/1\) Check the 1,1 cases first, to make the arithmetic easier.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did the synthetic division but Im not sure what to do after, dw:1360357822967:dw

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1I don't remember synthetic division, I always do it the long division way :) \[x^4 + 3x^3  5x^2 9x 2\]We can take out x^3(x+1)\[2x^35x^29x2\]We can take out 2x^2(x+1)\[7x^29x2\]We can take out 7x(x+1)\[2x2\]We can take out 2(x+1) so our quotient is \[x^3+2x^27x2\]Now you can repeat the whole process of picking a root, factoring it out, wash, rinse, repeat...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Long division, oh boy haha.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, so do I continue to take out x+1 from the rest?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1You could try, but do you think x=1 is a root of \[x^3 + 2x^2  7x  2 = 0\]If it isn't, you won't get far trying to divide out (x+1)...

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1well, it won't come evenly, at least.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Use the RRT to pick your next root to factor out.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Because you've got the same leading coefficient, and the same trailing coefficient, you've got the same list of tentative roots to try: 1, 1, 2, 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I dont really understand the rational root therom..

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Okay, here's an intuitive way to think about it: we know that if we have \[P(x) = (xa)(xb) = 0\]\[P(x) = x^2 ax  bx + ab = x^2 (a+b)x + ab\]Right? Similarly, if we had \[P(x) = (xa)(xb)(xc) = 0\]\[P(x) = x^3  (a+b+c)x^2 + (ab + ac + bc)x abc\]and so on. You can see that if the coefficient of the leading term is 1, then the coefficient of the last term is just the product of the various zeros, right?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1We may have more candidates than we do roots, of course, which is why we have to do some trial and error. For example, if abc = 1, and we know all the roots are integers, then our job is very easy: there's a root that = 1, and one that = 1, and another one that equals 1. Does any of it make sense to you? Maybe I can build on that...

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Are you comfortable with the notion that a polynomial with roots a, b, c, etc. can be written \[P(x) = (xa)(xb)(xc)\dots = 0\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, the a,b,c really confuses me

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Well, okay, let's say we have a polynomial with roots x = 1, and x = 2. That means that at x = 1, the polynomial = 0, and at x = 2, the polynomial = 0. Good so far?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1that's the definition of a root of a polynomial P(x) is that at the root, P(root) = 0.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1So for our polynomial with roots at x = 1, x = 2, P(1) = 0 and P(2) = 0.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, got it so far, I think..

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Good! Okay, there's a theorem that says if our polynomial has \(n\) roots, the highest order term in the polynomial will be \(x^n\) (and vice versa).

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1I don't know best how to explain some of this, but you can take it as a given that our polynomial with roots x = 1 and x =2 can be written\[P(x) = (x1)(x2) \]and it is easy to see that \(P(x) = 0\) if \(x = 1\) or \(x = 2\) right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay got it so far to here,

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1This is why if we can factor a polynomial into that form, we can find the roots easily by setting each of those product binomials \((x1) = 0\), \((x2) = 0\) and so on.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Basically, if we have some number of expressions multiplied together, and the product is 0, we know that one or more of those expressions must also be 0, and we can find all the ways to make the whole thing be 0 by finding the solution for each expression to = 0.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1So as long as we believe that we can write our polynomial in that form (this is the part I don't want to try to prove here), we're in good shape! So believe that, okay? :)

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Now back to the RRT and how we could pick roots to try from some big ugly polynomial that is multiplied out! We've got our little "test" polynomial with roots x = 1 and x = 2. \[P(x) = (x1)(x2)\]Let's multiply that out and see what we get:\[P(x) = (x1)(x2) = x^2  2x  1x + 2 = x^2  3x + 2\] Agreed?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Agreed, I need to run to the store right quick, will you be still be here in about an hour?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1Where did the value of 2 as the final coefficient come from? Well, it came exclusively from multiplying the roots! Let that sink in while you go to the store. I'll be back later on as well, but will continue writing even if you aren't here.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1What if we decide to add x = 3 as a root to our polynomial? We can multiply the whole thing by \((x+3)\): \[P(x) = (x1)(x2)(x+3) = (x+3)(x^23x+2) = \]\[P(x) = x^33x^2+2x+3x^29x+6\]\[P(x)= x^37x+6\] Again, the trailing coefficient comes entirely from the multiplication of the roots. Let's throw in one more root, x = 5: \[P(x) = (x1)(x2)(x+3)(x+5) = (x^37x+6)(x+5)\]\[P(x) = x^4+5x^37x^235x+6x+30\]\[P(x) = x^4+5x^37x^229x+30\] Once again, that final term (30) came from multiplying our roots together. Let's say we found that \(P(x)\) on our homework assignment, and were told to find the roots? I don't know about you, but I can't just guess numbers that will work (well, okay, in this case I would try 1 as a guess, and it would work, but in general...) So we could use the RRT to get some numbers to try. What are the factors of 30? 1, 2, 3, 5, 6, 10, 15, 30 (and the negative numbers as well). We know that the zeros are on that list, because that final factor came just from multiplying the zeros together. Unfortunately, there are more values on the list than there are zeros :)

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1But I hope you can see at this point that if we can get 1 root of the big polynomial, we can divide out (x  root) from the whole thing and get a polynomial of a lower degree, at which point we can repeat the process over and over until we have found all of them.

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1And as we divide out (xroot) the number of possible factors of the final term gets smaller, at least if the root is some number other than 1 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Makes a little more sense, still confused on finding the zeros? Thank you for taking your time out & explaining this to me.! :)
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