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List all possible rational solutions:
f(x)=2x^(3)-5x+7
asked Jan 11, 2013 in TRIGONOMETRY by angel12 Scholar

2 Answers

0 votes

f (x) = 2(x^3) - 5x+7 it is wrong

f (x) = 2(x^3) +5x +7

This equation it can be written as

f (x) = 2(x^3) +0x^2 +5x +7

Synthetic Division Method

      -1  |      2       0         5       7

           |      0      -2       2       -7

           |______________________

                 2       -2        2       0

Now it can be written as this equations is

f (x) = (x+1)(2(x^2)-2x+2)

Now slove the equation factor method

Consider

= (2(x^2)-2x+2)                     2*2 = 4  ,  -4 + 2 = - 2

= 2(x^2)-4x+2x+2

Take out common factors.

= 2x(x-2) +2 (x-2)

Take out common factors.

= (x-2)(2x-2)

There fore

f (x) = (x-1)(x-2)(2x-2)

If f (x) = 0

Than

(x-1)(x-2)(2x-2) = 0

x-1 = 0  or  x-2 = 0  or  2x-2 = 0

x = 1  or  x = 2 or 1 = 1

There fore x = 1 , 2 , and 1

answered Jan 12, 2013 by richardson Scholar
0 votes

The function is f(x ) = 2(x3 ) + 5x + 7.

If p/q is a rational zero, then p is a factor of 7 and q is a factor of 2.

The possible values of p are   ± 1,   ± 7.

The possible values for q are ± 1 and   ± 2.

So, p/q = ± 1,   ± 7,   ±1/2,   ± 1/7

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

p/q

2

0

5

7

1

2

2

-1

6

-1

2

-2

7

0

Since f(-1)   =   0,   x  =  -1 is a zero. x + 1 is a root.

The depressed polynomial is  2x– 2x + 7 = 0.

Since the depressed polynomial of this zero, 2x– 2x + 7, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation.

2x– 2x + 7 = 0

x = {-b ± √ (b2 - 4ac)} / 2a where a = 2, b = -2 and c =7

x = {-(-2) ± √ ((-2)2 - 4(2)(7))} / 2(2)

x = {2 ± √ (4 - 56)} / 4

x = {2 ± √(-52)} / 4

x = {2 ± i√(4 * 13)} / 4

x = {2 ± 2i√(13)} / 4

x = {2( 1 ± i√13 )} / 4

x = (1 ± i√13 ) / 2

Therefore x = (1 + i√13 ) / 2, (1 - i√13 ) / 2.

Therefore possible solutions are x = -1, (1 + i√13 ) / 2, (1 - i√13 ) / 2.

answered Jun 30, 2014 by joly Scholar

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