1. ## Solve x

Solve each of the following equations for x:

$12x^{4}+11x^{3}-26x^{2}+x+2=0$
$x^{4}+2x^{3}-3x^{2}-4x+4=0$
$6x^{4}-5x^{3}-20x^{2}+25x-6=0$

If possible, any help with step-by-step working out would be highly appreciated.

2. Notice that $x=1$ satisfies all the three equations. So $x=1$ is a root for each of them. Consequently, $(x-1)$ is a factor of each of them. Dividing each by $(x-1)$ we have:

$-2-3x+23x^2+12x^3=0$
$-4+3x^2+x^3=0$
$6-19 x+x^2+6 x^3=0$

For the second equation, $x=1$ further gives a solution; so divide it again by $(x-1)$. Thus your job reduces to solve $4+4x+x^2=0$

For the first and third equation, and the reduced form of the second equation, try to factorize. This gives for the first equation and reduced second equation
$(x+2) (3 x-1) (4 x+1)=0$
$(x+2)^2=0$ respectively; so the solutions are $x=-2, \frac{1}{3} , -\frac{1}{4}$ and $-2, -2$ respectively.

For the third one, put values of $x$ and see where $f(x)$ changes sign. Once you find two such consecutive values, you get one root between those x-values. Prceed accordingly.

3. Originally Posted by Sambit
Notice that $x=1$ satisfies all the three equations. So $x=1$ is a root for each of them. Consequently, $(x-1)$ is a factor of each of them. Dividing each by $(x-1)$ we have:

$-2-3x+23x^2+12x^3=0$
$-4+3x^2+x^3=0$
$6-19 x+x^2+6 x^3=0$

How do you factorise the first and third equation?
Could you please show all steps as I still do not understand.

How did you get:
Originally Posted by Sambit
$(x+2) (3 x-1) (4 x+1)=0$

from:
Originally Posted by Sambit
$x=1$ $-2-3x+23x^2+12x^3=0$
?

4. Originally Posted by Joker37
Solve each of the following equations for x:

$12x^{4}+11x^{3}-26x^{2}+x+2=0$
$x^{4}+2x^{3}-3x^{2}-4x+4=0$
$6x^{4}-5x^{3}-20x^{2}+25x-6=0$

If possible, any help with step-by-step working out would be highly appreciated.
The growth is in the exploration, not necessarily in the solution.

1) Rules of Signs - This is an existence idea.

Line up the exponents in the usual decreasing order. Count the sign changes. I get 2, 2, 3. This is the maximum number of positive Real zeros.

Change the sign on the odd exponents and do it again. I get 2, 2, 1. This is the maximum number of negative zeros.

The exciting part is that if we don't get the maximum, it must be different by a factor of 2. This means the third MUST have two real zeros, one positive and one negative. Find a section on Complex Conjugates to get a handle on this.

2) Rational Root Theorem

If there are rational zero, they must be related to the leading and trainling coefficients. I get these as possibilities.

1, 1/2, 1/3, 1/4, 1/6, 1/12, 2, 2/3, both positive and negative.
1, 2, 4, both positive and negative
1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 6, both positive and negative

While you are checking these, you can also check for maxima and minima. For example, assuming you are checking with synthetic division, if you get all positive values across the bottom, please don't bother to try anything greater than whatever led to that result.

3) Mean Value Theorem

While you are checking rational roots, keep track of your results. If you don't find any rational zeros, the signs of the results may be beneficial in tracking down irrational zeros.

4) When you find a zero, particularly a rational one, factor your current polynomial and start over with the new one of one degree less. Again, you can use the bottom row of your synthetic division to factor quickly, since it's already done for you.

I worry about request for step-by-step. You probably already have a text book. Will it really be of any benefit to write another one? Let's see what you get.

5. Originally Posted by Joker37
Solve each of the following equations for x:

$12x^{4}+11x^{3}-26x^{2}+x+2=0$
Hi there, Sambit has given you some good advice here, he/she has applied the facotr theorem which is:

For $f(x)$ , if $f(a) = 0$ then $x-a$ is a factor.

You were told in post #2 that $x-1$ is a factor, is it clear why?

Then we divide, $(12x^{4}+11x^{3}-26x^{2}+x+2) \div (x-1)$

What is the result? This is also given to you in post #2