f(x)=2x^4-17x^3+59x^2-83x+39

Question

I have to find the zeros

Answer 1

If we add all the coefficients we get 2-17+59-83+39=100-100=0, so x=1 is a zero.

Divide by this zero using synthetic division:

1 | 2 -17   59 -83    39

     2    2 -15   44 | -39

     2 -15  44  -39 |    0 = 2x³-15x²+44x-39 = g(x)

Add the coefficients of g(x): 2-15+44-39=46-54=-8, so x=1 is not another zero, neither is -1, because all the terms would be negative. The factors of 39 other than 1 and 39 are 3 and 13, and the factors of the x³ term are only 2 and 1. So we could have zeroes ½, 1½, 6½, 3, 13, 19½. Another clue is to plug in x=2 and see if there's a change of sign:

16-60+88-39=104-99=5. So g(1)=-8 and g(2)=5, therefore there is a zero between x=1 and x=2. It's possible that x=1½=3/2, so let's try it:

2(27/8)-15(9/4)+44(3/2)-39=(27-135+264-156)/4=(291-291)/4=0, therefore x=1½ is a zero. Divide by this zero using synthetic division:

1½ | 2 -15   44   -39

        2    3  -18 |  39 

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

f(x)=(x-3/2)(x-1)(2x²-12x+26)=(2x-3)(x-1)(x²-6x+13).

Therefore the real zeroes are 1 and 1½ (1.5).

If we solve the quadratic x²-6x+13=0 we get the complex zeroes:

x²-6x+9-9+13=0, (x-3)²=-4, x-3=±i√4, giving conjugate zeroes 3+2i and 3-2i.