finding the real zeros of a Polynomial

Question

p(x) = 4x^3-6x^2+1

Answer 1

First, we list all the possible zeros, c, of p(x).  Take the factors of the last term, r, divided by the factors of the first term, q.

 

c = ± r / q

 

c = ±(1, 1/2 , 1/4)

 

 

Next, we use synthetic division to determine which of these is a zero.  Then, once we find the zero, put it into factor form.

Synthetic division divides the coefficients of p(x) by the possible zeros, c.  The remainder must be zero.

 

 

Trying 1/2 as a possible zero,

 

 

(1/2)  |   4      -6        0         1

 

                       2       -2       -1

__________________________

              4      -4       -2         0

 

 

The digits under the line are the coefficients of the quotient.  The last digit is the remainder.  Since the last digit is zero, and needs to be zero, we found our factor.

 

p(x) = (2x - 1)(4x² - 4x - 2)

 

p(x) = (2x - 1) * 2(2x² - 2x - 1)

 

p(x) = 2(2x - 1)(2x² - 2x - 1)

 

 

Set p(x) equal to zero.

 

0 = 2(2x - 1)(2x² - 2x - 1)

 

 

Now, use the quadratic formula to find the other zeros from the larger factor.  From the formula, we can find the discriminant  to see if we get a positive or negative value.  If negative, then the zeros are complex.  If positive, then the zeros are real.

 

b² - 4ac = (-2)² - 4(2)(-1)

            = 4 + 8

            = 12

 

 

Value is real, so we continue working.

 

So from the quadratic formula,  

 

x = (2 ± √12) / 4

 

x = (2 ± 2√3) / 4

 

x = (1 - √3) / 2              and              x = (1 + √3) / 2