6x^3 + 5x^2-6x-5=0?

+1 vote

6x3+5x2-6x-5=0

this eqution it can be written as

(x+1)(6x2+11x+5)=0

now we can slove the equation factor method

(x+1)(6x2+6x+5x+5)=0                                   (6*5=30 , 6+5=11)

Take out common factors.

(x+1)[6x(x+1)+5(x+1)]=0

Take out common factors.

(x+1)(x+1)(6x+5)=0

than (x+1)=0, or  (x+1)=0, or  (6x+5)=0

x=-1  or     x=-1    or      6x+5=0

Subtract 5 from each side.

6x+5-5=0-5

6x=-5

Divide each side by 6

(6x)/6=-5/6

x=-5/6

there fore x=-1,x=-1,x=-5/6

x= -1, -1, -5/6

+1 vote

6x3+5x2-6x-5=0

|    6   5   -6   -5

|   0   6   11    5

-------------------------

6   11   5   0

this eqution it can be written as

(x- 1)(6x2+11x+5)=0

now we can slove the equation factor method

(x-1)(6x2+6x+5x+5)=0                                   (6*5=30 , 6+5=11)

Take out common factors.

(x-1)[6x(x+1)+5(x+1)]=0

Take out common factors.

(x-1)(x+1)(6x+5)=0

than (x-1)=0, or  (x+1)=0, or  (6x+5)=0

x=1  or     x=-1    or      6x+5=0

Subtract 5 from each side.

6x+5-5=0-5

6x=-5

Divide each side by 6

(6x)/6=-5/6

x=-5/6

there fore x=1,x=-1,x=-5/6

+1 vote

6x3+5x2-6x-5=0

Synthetic method

-1 |  6     5      -6      -5

|  0    -6      1        5

|________________________

6      -1      -5      0

Now it can be written as

(x+1)(6x2-1x-5)=0

now we can slove the equation factor method

(x+1)(6x2-6x+5x-5)=0                                   (-6*5=-30 , -6+5=-1)

Take out common factors.

(x+1)[6x(x-1)+5(x-1)]=0

Take out common factors.

(x+1)(x-1)(6x+5)=0

than (x+1)=0, or  (x-1)=0, or  (6x+5)=0

x=-1  or     x=1    or      6x+5=0

Subtract 5 from each side.

6x+5-5=0-5

6x=-5

Divide each side by 6

(6x)/6=-5/6

x=-5/6

there fore x=-1,x=1,x=-5/6

x= -1, 1, -5/6

Identify Rational Zeros

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

6x3+5x2-6x-5 = 0

If p/q is a rational zero, then p is a factor of 5 and q is a factor of 6.

The possible values of p are   ± 1 and   ± 5.

The possible values for q are ± 1, ± 2, ± 3,± 6.

So, p/q = ± 1, ± 1/2, ± 1/3, ± 1/6,± 5, ± 5/2, ± 5/3, ± 5/6.

Make a table for the synthetic division and test possible  zeros.

 p/q 6 5 -6 -5 1 6 11 5 0 2 6 17 28 51 -1 6 -1 -5 0

Since f(-1) = 0, x  = -1 is a zero.

Therefore (x + 1) is a root.

The depressed polynomial is 6x2 - x - 5 = 0

Since the depressed polynomial of this zero, 6x2 - x - 5 , is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation or find factors if possible.

6x2 - x - 5 = 0

6x2 - 6 x + 5x - 5 = 0

6x(x - 1) + 5(x - 1) = 0

(6x + 5)(x - 1) = 0

(6x + 5) = 0 or (x - 1) = 0

Take 6x + 5 = 0

6x = - 5

x = - 5/6

Therefore (x + 5/6) is a root.

Take (x - 1) = 0

x - 1 = 0

x = 1

Therefore (x - 1) is a root.

Therefore the roots are (x + 1) (x + 5/6) (x - 1).