2x^4-19x^3+71x^2-109x+39=0 how do i work this step by step

Question

i can find the roots i see that 2 is my p 39 is my Q i can synthetically divide by 3 and then by 1/2 i come up with the ending of 2x^2-12x+26 i know from there i can use the quadratic formula after that i do not know how to complete it. please help

Answer 1

First check out the rational zeroes by looking at the factors of the coefficients of the highest and lowest degrees of the polynomial, that is, factors of 2 (coefficient of x⁴) and factors of 39 (constant).

39 has 1, 39, 3 and 13 as factors; 2 has only 1 and 2.

Rational zeroes are:

1, 3, 13, 39, 1/2, 3/2, 13/2, 39/2 (positive and negative forms).

So there are 16 rational zeroes. We set x to each of these values and evaluate the polynomial. If we find any values that evaluate to zero, these are true zeroes (roots). To cut a long story short we can identify 1/2 and 3 as true zeroes. Using synthetic division we can reduce the polynomial to a quadratic.

3 | 2 -19  71 -109    39

     2    6 -39    96 | -39

½ |2 -13  32   -13 |    0 

     2    1   -6 |  13

     2 -12  26 |    0 = 2(x²-6x+13) which has no real zeroes.

The common factor 2 in this quadratic can be used to convert factor x-½ into 2x-1.

So complete factorisation is (2x-1)(x-3)(x²-6x+13).

The quadratic has a complex solution:

x²-6x=-13,

x²-6x+9=-13+9=-4,

(x-3)²=-4, x-3=±2i, so x=3+2i or 3-2i.

The zeroes are: ½, 3, 3+2i, 3-2i.