Polynomials

Question

Given the polynomial p(x)=2x3-5x2-6x+4
a. Factor the polynomial completely into linear factors
b. List all the zeros of p(x) (in exact form)

Please explain how you got the answers
Thanks!

Answer 1

Using the Rational Root Theorem and Synthetic Division,

one root is 0.5.

One factor is (x - 0.5)

the other factor is 2(x² - 2x - 4)

Use the Quadratic Formula to determine the other two roots.

Answer 2

p(x) = 2x³-5x²-6x+4

From the rational root theorem the real root must be

contained within the factors of the constant (4) divided

by the factors of the coefficient of the highest term (2)..

So the factors of 4, over the factors of 2

 

The real root is within:

±(1, 2, 4)

-----------   = ±(1, 1/2, 2, 4)

  ±(1, 2)

 

Use synthetic division to determine a root:

 

If x=1 is a root then (x-1) is a factor

 

1 | 2    -5    -6   4

   |        2    -3  -9

   -------------------

     2    -3    -9  -5

Since the remainder is -5, (x-1) is not root

*******************************

If x=-1 is a root then (x+1) is a factor

 

-1 |  2     -5     -6    4

    |         -2      7   -1

    ------------------------

      2      -7      1     3

Since the remainder is 3, this is not a root

******************************

If x = 1/2 is a root then (x-1/2) is a factor

 

1/2 |    2     -5     -6    4

      |            1     -2   -4

      ------------------------

           2     -4     -8    0

 

The remainder here is zero so x = 1/2 is a root

of p(x).  Factoring out (x-1/2) leaves

2x²-4x-8   as can be seen from the synthetic division.

 

p(x) = (x-1/2)(2x²-4x-8)

 

Note that a 2 can be factored out of 2x²-4x-8 so:

 

p(x) = 2(x-1/2)(x²-2x-4)

 

We need to use the quadratic formula to find the

roots of x²-2x-4

 

x = [-b ±√(b²-4ac)]/2a    a=1, b=-2, c=-4

 

x = [2 ±√((-2)²-4(1)(-4))]/2(1)

x = [2 ±√(20)/2

x = [2 ±2√5]/2

x = 1 ±√5

 

The 3 roots are x =1/2, 1+√5, 1-√5

 

The polynomial factors completely out to:

 

p(x) = 2(x-1/2)(x-1-√5)(x-1+√5)