SOLUTION: Solve 6x⁴ - 35x³ + 62x² - 35x + 6 = 0

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Question 1131704: Solve 6x⁴ - 35x³ + 62x² - 35x + 6 = 0
Found 4 solutions by rothauserc, MathLover1, ikleyn, MathTherapy:
Answer by rothauserc(4718) About Me  (Show Source):
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6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0
:
check for the factors of 6(this is the constant and also the leading coefficient in the above equation)
:
use the Rational Zeros Theorem
:
factors for 6 are 1, 2, 3, 6
:
the possible zeros are
:
(+ or - 1, 2, 3, 6)/(1, 2, 3, 6) = + or - 1, 2, 3, 6, 1/2, 1, 3/2, 3, 1/3, 2/3, 1, 2, 1/6, 1/3, 1/2, 1 =
:
-6, -3, -2, -3/2, -1, -1/2, -2/3, -1/3, -1/6, 1/6, 1/3, 2/3, 1/2, 1, 3/2, 2, 3, 6
:
look at the graph to see if we can eliminate any of the possible zeros
:
+graph%28+300%2C+200%2C+-1%2C+4%2C+-10%2C+50%2C+6x%5E4+-+35x%5E3+%2B+62x%5E2+-+35x+%2B+6%29+
:
from the graph we see that there are no negative roots and 4 positive roots
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we can eliminate 1/6, 2/3, 1, 3/2, 6
:
*******************************************************
the roots are 1/3, 1/2, 2, 3
:
to check this substitute for x in the equation
*******************************************************

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Solve
6x%5E4+-35x%5E3+%2B+62x%5E2+-35x+%2B+6+=+0+....factor completely
write -35x%5E3 as -2x%5E3-3x%5E3-12x%5E3-18x%5E3
and 62x%5E2 as %2Bx%5E2%2B4x%5E2%2B6x%5E2%2B6x%5E2%2B9x%5E2%2B36x%5E2

...group




%28x-3%29%286x%5E3-2x%5E2-3x%5E2-12x%5E2%2B4x-2%2Bx%2B6x%29
%28x-3%29%28%286x%5E3-12x%5E2%29-%282x%5E2-4x%29-%283x%5E2-6x%29%2B%28x-2%29%29
%28x-3%29%286x%5E2%28x-2%29-2x%28x-2%29-3x%28x-2%29%2B%28x-2%29%29
%28x-3%29%28x-2%29%286x%5E2-2x-3x%2B1%29
%28x-3%29%28x-2%29%28%286x%5E2-3x%29-%282x-1%29%29
%28x-3%29%28x-2%29%283x%282x-1%29-%282x-1%29%29
%28x+-+3%29+%28x+-+2%29+%282x+-+1%29+%283+x+-+1%29+=+0

solutions:
x=3
x=2
x=1%2F2
x=1%2F3

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Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.

            This equation of the degree  4  is VERY SPECIAL.
            It relates to the class of so named  palindromic  equations  of the degree  4,  which means that its coefficients
            form a palindromic sequence.

            See this Wikipedia article https://en.wikipedia.org/wiki/Reciprocal_polynomial#Palindromic_polynomial

            There is a  SPECIAL  PROCEDURE  in algebra to solve such equations.  It is presented below.


6x%5E4+-+35x%5E3+%2B+62x%5E2+-+35x+%2B+6 = 0      (1)


It follows from the equation that x= 0 IS NOT the root.
So, we can divide both sides by  x%5E2  without loosing the roots. In this way, you will get an equivalent equation


6x%5E2+-+35x+%2B+62+-+35%2A%281%2Fx%29+%2B+6%2A%281%2Fx%5E2%29 = 0.


Group and re-write it equivalently in the form


6%2A%28x%5E2+%2B+2+%2B+1%2Fx%5E2%29 - %2835x+%2B+35%2A%281%2Fx%29%29 + 50 = 0,    or


6%2A%28x%2B1%2Fx%29%5E2 - 35%2A%28x%2B1%2Fx%29 + 50 = 0.     (2)


Introduce new variable  u = x + 1%2Fx.  Then equation (2) takes a form


6u%5E2+-+35u+%2B+50 = 0.


Solve this quadratic equation using the quadratic formula


u%5B1%2C2%5D = 35+%2B-+sqrt%2835%5E2+-+4%2A6%2A50%29%29%2F%282%2A6%29 = %2835+%2B-+sqrt%2825%29%29%2F12.


The two roots are  

    u%5B1%5D = %2835+%2B+sqrt%2825%29%29%2F12 = %2835+%2B+5%29%2F12 = 40%2F12 = 10%2F3  and

    u%5B2%5D = %2835+-+sqrt%2825%29%29%2F12 = %2835+-+5%29%2F12 = 30%2F12 = 10%2F4.


Now, to find x,  we need to solve two equations

    a)  x + 1%2Fx = 10%2F3   and  b)  x + 1%2Fx = 10%2F4.



Case a).  x + 1%2Fx = 10%2F3

          3x%5E2+-+10x+%2B+3 = 0

          x%5B1%2C2%5D = %2810+%2B-+sqrt%28%28-10%29%5E2+-+4%2A3%2A3%29%29%2F%282%2A3%29 = %2810+%2B-+sqrt%2864%29%29%2F6 = %2810+%2B-+8%29%2F6.

          So, the two roots are  x%5B1%5D = %2810+%2B+8%29%2F6 = 3  and  x%5B2%5D = %2810+-+8%29%2F6 = 1%2F3.



Case b).  x + 1%2Fx = 10%2F4

          4x%5E2+-+10x+%2B+4 = 0

          x%5B3%2C4%5D = %2810+%2B-+sqrt%28%28-10%29%5E2+-+4%2A4%2A4%29%29%2F%282%2A4%29 = %2810+%2B-+sqrt%2836%29%29%2F6 = %2810+%2B-+6%29%2F8.

          So, the two roots are  x%5B3%5D = %2810+%2B+6%29%2F8 = 2  and  x%5B4%5D = %2810+-+6%29%2F8 = 1%2F2.


ANSWER.  The four roots are   1%2F3,  1%2F2,  2  and  3.

Solved.


The lesson to learn

    From this post learn on how to solve palindromic equations.

    Every palindromic equation of the degree 4 can be solved in this way.


The major steps of the solution are :

    a)  Divide both sides by  x%5E2;

    b)  Introduce new variable  u = x + 1%2Fx;

    c)  Reduce the equation to a quadratic equation relative new variable u  and solve it getting two roots  u%5B1%5D  and  u%5B2%5D;

    d)  Then find  x  by solving two equations  x%2B1%2Fx = u%5B1%5D  and  x%2B1%2Fx = u%5B2%5D.

Again :
    Every palindromic equation of the degree 4 can be solved in this way.


Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Solve 6x⁴ - 35x³ + 62x² - 35x + 6 = 0
RATIONAL ROOT THEOREM produces 2 zeroes: 2 and 3. Therefore, 2 of the expression's factors are: (x - 2) and (x - 3). 
FOILing these 2 factors results in trinomial: x2 - 5x + 6
Now, when 6x4 - 35x3 + 62x2 - 35x + 6 is divided by x2 - 5x + 6, we get: 6x2 - 5x + 1. Factoring 6x2 - 5x + 1 gives us: (3x - 1)(2x - 1).
Therefore, 6x4 - 35x3 + 62x2 - 35x + 6 = 0 becomes: (3x - 1)(2x - 1)(x - 2)(x - 3) = 0 and the solutions are: