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Find all zeros of the polynomial 6x4+17x3-2x2+x-6 = 0

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giokarinos | Student, Undergraduate | eNoter

Posted January 22, 2011 at 5:59 PM via web

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Find all zeros of the polynomial 6x4+17x3-2x2+x-6 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 23, 2011 at 1:36 AM (Answer #1)

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We can write the given polynomial 6*x^4+17*x^3-2*x^2+x-6 as:

6*x^4+17*x^3-2*x^2+x-6

=> 6x^4 + 18x^3 - x^3 - 3x^2 + x^2 + 3x - 2x - 6

=> 6x^3( x + 3) - x^2(x + 3) + x(x + 3) - 2(x + 3)

=> (x + 3)[6x^3 - x^2 + x - 2]

=> (x + 3)( 6x^3 - 4x^2 + 3x^2 - 2x + 3x - 2)

=> (x +3)(2x^2(3x - 2) + x(3x - 2) + 1(3x - 2))

=> (x +3)(3x - 2)( 2x^2 + x +1)

Equating the polynomial to 0

(x +3)(3x - 2)( 2x^2 + x +1)

x + 3 = 0

=> x1 = -3

3x - 2 = 0

=> x2 = 2/3

2x^2 + x +1 = 0

=> x3 = [-1 + sqrt (1 - 8)]/4

=> x3 = -1/4 + (i*sqrt 7) / 4

x4 = -1/4 - (i*sqrt 7) / 4

Therefore the roots or zeroes of the polynomial 6*x^4+17*x^3-2*x^2+x-6 are :

-3 , 2/3 , [(-1/4) + (i*sqrt 7)/4] , [(-1/4) - (i*sqrt 7)/4]

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academicsfirst | High School Teacher | (Level 2) Adjunct Educator

Posted January 23, 2011 at 1:32 AM (Answer #2)

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We can find the zeros of a polynomial equation by using synthetic division.

First check to see that the equation is written in descending powers of the variable.

Then look at the highest power in the equation.  This will tell how many solutions (zeros) there are for the equation.

6x^4   +  17x^3  -  2x^2  +  x  -  6   =   0

The degree of the equation is 4; therefore, we will have 4 solutions.

Synthetic Division:

Step 1:    Determine possible zeros by using (P)/(Q).  The values for (P) will be from the factors of the last term which is -6.  The values for (Q) will be from the factors of the coefficient of the first term, which is 6.

Possible zeros:   ±1,±2,±3,±6,±½,±⅓,±1/6

 

Step 2:    Write the coefficients of the equation.  The goal of testing a possible zero value is to arrive at a remainder of zero in the last column of your computations.   It will help to go in order of the possible zeros beginning with 1 and continuing until you have found a zero by the following process. By trial and error, I have already determined that -3 and 2/3 are zeros for this polynomial.  The following computations were used to identify the two solutions.

Step 1:    Write the coefficients of the terms.

Step 2:    Write the test zero to the far left of the second column.

Step 3:    “Bring down” the first coefficient

Step 4:   Multiply the first coefficient by the test value (6 * -3) and place the product (-18) in the second column as shown.  Now, find the sum of the two numbers (-1) and write it directly underneath the -18.  Multiply the -1 by the -3 and write the product (3)  under the -2 in the third column.  Find the sum (1).  Continue the pattern until the sum of the two digits in the last column is 0.  This verifies that -3 is a remainder.

6        17        -2        1        -6

-3*                                  -18          3        -3        6

6         -1           1        -2        0

Using the third row as new coefficients and dropping the degree of the equation by one our new “depressed” equation is

6x^3 -  x^2  +  x   -   2   =  0

Use the coefficients from the new equation to continue with the next test value as shown below.

 

6          -1          1         -2

2/3*                            4           2          2

6           3          3          0

We have now depressed the equation twice.  Our new equation is a quadratic.

6x^2  +  3x  +  3   =  0

We can simplify the equation by factoring out  3.

3(2x^2  +  x  +  1) = 0

Dividing by 3 on both sides, our simplified equation is

2x^2  +  x  +  1  =  0

We will use the quadratic formula to find the other 2 roots.

-b±sqr rt (b²-4ac) / 2a

From the coefficients of our quadratic equation we know the following:  a=2; b=1; c=1

-1±sqr rt(1²-4(2)(1)) / 2(2)

After simplifying, our answer is  ( -1±√-7)/4

However, since we cannot have a negative square root we extract the imaginary “i”

The two roots from the quadratic  are    (-1±i√7)/4

The  four  zeros for our original polynomial equation are as follows:  (-3,   2/3,  (-1±i√7)/4 )

 

 

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