# Thread: Polynomials - Solving Equations

1. ## Polynomials - Solving Equations

I've done quite a lot of practice with this topic but I'm stuck at these kind of questions.

Q) Find all the real roots of the equation $x^5-3x^4+2x^3-2x^2+3x+1=2x$

I've solved one root which is 1, but I got that by using the guess factor and putting in different values of x to equate the equation.

Can you show me the easiest way of solving these kind of questions.

Thanks in advance!

2. Quintics are dicey. In fact, there is a proof that although there are formulas for the general quadratic, cubic, and quartic polynomials, there is no general formula that solves the quintic. You can only solve special cases. In this case, 1 and -1 would be the numbers to try, because of the rational root theorem. After trying those, if you can reduce the order by factoring, you might be able to use the cubic formula or the quartic method.

3. Ackbeet mentioned the rational roots theorem, which gives us 1 and -1 as possibilities. Divide the original polynomial by x - 1 using synthetic division and see what you get.

4. Well this is a High School level question, so I am assuming there must be a simpler method to solve this? How can I factorize this?

Better yet can someone show me an easy method of solving such questions.

Thanks!

5. $x^5-3x^4+2x^3-2x^2+x+1=0$

real roots are $1, 1\pm\sqrt{2}$

6. Well that is correct, but how can I solve this myself. Can you please show me the method.
Thanks!

7. Originally Posted by unstopabl3
Well that is correct, but how can I solve this myself. Can you please show me the method.
Thanks!
Ackbeet's post told you how to find the real root x = 1.

x = 1 is a root => x - 1 is a factor. Divide that factor into your quintic using polynomial long division and you get $x^4 - 2x^3 -2x-1$ as another factor.

Your new job now is to solve $x^4 - 2x^3 -2x-1 = 0$.

Note that the left hand side of the above can be written $x^4 - 2x^3 - x^2 + x^2 - 2x - 1 = x^2(x^2 - 2x - 1) + ( x^2 - 2x - 1)$ ....

8. well what is left from the first factor is
$(x^2+1)(x^2-2x-1)$
to find factors of $(x^2-2x-1) \rightarrow (x-1)^2 -2 \rightarrow x= 1\pm\sqrt{2}$
this was done by completing the square.

do you need more steps.
i don't think there is a slam dunk to this one.

EDIT: clarified step

9. $x^5 - 3x^4 + 2x^3 - 2x^2 + x + 1=0$
As I said, since you know that one of the roots is 1, use synthetic division:
Code:
1|  1 -3  2 -2  1  1
--     1 -2  0 -2 -1
---------------------
1 -2  0 -2 -1  0
So your equation is the same as
$(x - 1)(x^4 - 2x^3 - 2x - 1) = 0$

Take the 2nd factor and solve for x:
$x^4 - 2x^3 - 2x - 1 = 0$

Subtract and add an x^2 on the left side:
$x^4 - 2x^3 - x^2 + x^2 - 2x - 1 = 0$

Factor by grouping three terms -- find the GCF for the first 3 terms, and do the same for the last 3 terms:
$x^2(x^2 - 2x - 1) + 1(x^2 - 2x - 1) = 0$

So this factors to
$(x^2 + 1)(x^2 - 2x - 1) = 0$

Set each factor to 0. For the first, there will be no real solutions. For the 2nd, use the quadratic formula.

EDIT: Ack! Too slow!