find the solution set for 2x^4-15x^3+32x^2-30x+56

Question

what is the solution set?

Answer 1

Find rational zeroes first:

Take factors of coefficients of the highest and lowest degree:

Factors of 2 are 1 and 2; factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56.

Rational zeroes are:

1, 2, 4, 7, 8, 14, 28, 56, 7, 7/2 (positive and negative forms).

Plug in these one by one:

x=2: 32-120+128-60+56=36;

x=4: 512-960+512-120+56=0. So x=4 is a zero.

Use synthetic division to reduce the degree of the polynomial:

4 | 2 -15  32 -30    56

     2    8 -28  16 | -56

     2   -7    4 -14 |    0

So now we have the cubic: 2x³-7x²+4x-14.

The rational zeroes are:

1, 2, 7, 14, ½, 7/2. We can see that ±1 is not a zero and we have already tried 2, and -2 would make the polynomial negative. 7 and -7 are not zeroes. 14 is not a zero, because 14³ is too large, just as 7 was.

x=½: ¼-7/4+2-14≠0;

x=7/2: 343/4-343/4+14-14=0; so 7/2 is another zero. Divide by this zero:

7/2 | 2 -7 4  -14

       2   7 0 | 14

       2   0 4 |   0

The quadratic is 2x²+4=2(x²+2). The distributed 2 converts the zero from x-7/2 to 2x-7 as a factor.

Therefore we have factorised the polynomial=(x-4)(2x-7)(x²+2). The two complex solutions are x=i√2 and -i√2.