POLYNOMIALS.... Please HELP ME?

Question

Okay, here's the question.

http://img373.imageshack.us/img373/6560/poy9vp3.png

I've been trying to solve this, but I'm a bit stuck. Can someone help me?

Answer 1

y = 4x^4 - 10x^3 -2x^2 -10x -6

if a/b is a rational root then ax-b is a factor.
leading coeff = 4 divisible by 4,2,1
trailing coeff = 6 divisible by 6,3,2,1

so the possible rational roots are
+/- 6,3,2,1, 3/2,1/2, 3/4,1/4

I can spot that -1/2 is a root:
y = 4/16 + 10/8 -2/4 +10/2 -6 = (2+10-4+40-48)/8 = 0
so we can factor out (2x+1)
y = 4x^4 - 10x^3 -2x^2 -10x -6
= (2x+1)(2x^3-6x^2+2x-6)
so the other possible rational roots are
+/- 6,3,2,1, 3/2,1/2,

I can spot that 3 is a root
y = (7)(2*27-6*9+2*3-6) =0
so we can factor out (x-3)

y=(2x+1)(x-3)(2x^2 +2)
the last is 2(x-i)(x+i) so there are no more rational roots

x1= -1/2, x2 = 3

Answer 2

Tom's solution is not correct and neither is Tanyeesern's.
Let's look at
4x^4 -10x³ - 2x²-10x-6 = 0.
First, divide out a 2, to simplify it a bit.
2x^4 - 5x^3 - x^2 -5x -3 =0. (*)
Now use the rational root theorem:
If a/b is a rational root of (*) then a divides 3
and b divides 2.
Testing all the possibilities we find that x = 3 is
a root, so x-3 is a factor of the LHS of (*)
Check 2*3^4 -5*3³ - 9 -15 -6 =0.
Dividing this synthetically or by long division,
the quotient is
2x³+x²+2x+1.
Setting this to zero and solving, we get
x²(2x+1) + (2x+1) = 0.
or
(x²+1)(2x+1) = 0.
So the other roots of (*) are
x = -1/2, x = i and x = -i.
The 2 rational roots are -1/2 and 3.
Hope that helped!

Answer 3

Start with the Rational Root Theorem to limit the number of guesses you have to make. When you find a zero, use it in a factor to divide the polynomial and create another reduced polynomial with a degree one less than the original. Keep repeating and eventually you'll reach a quadratic equation you can solve by factoring. I'm confused too, but my second link has pretty good step by step instructions. Solving higher order polynomials is really just a sophisticated form of guessing and trial and error.

Answer 4

First, I graphed the function and noticed that it intersected the x-axis at -0.5 and 3.. so this gave me two of the roots.

Then I used synthetic division to divide the polynomial by (x + 1/2) and then the result by (x - 3) and I got 4x^2 + 4 = 0

So this means I have two imaginary roots (since this can not be factored). If I solve this equation with the quadratic formula, I get i and -i

So the real roots are 3 and -1/2
the imaginary roots are i and -i

Answer 5

There is a theorem that for any polynomial f[x] with coefficients integers, all its rational zeros are integers.

It can be checked that the differential of the function with respect to x( assuming doing it in real numbers) is positive for all x>3 and negative for all x<0.

Since f(3)=0 and f(-1)>0, it can be concluded that there are no solutions for x>3 and x<-1 in reals, and so in rationals.

Only thing needed to check is to evaluate x at 0,1,2, they are all not zero.

Conclusion: there is only a zero, 3.

Answer 6

I'm not going to do it all because solving quartics is really boring.

But one solution is x=-1

Which means that (x+1) is a factor of that equation.

So you can divide through by that factor using polynomial division and find a cubic, which you can set about solving in a similar way.