how to solve h(x) = 66x^6 - 60x^5 + 137x^4 - 130x^3 - 101x^2 + 80x + 8

Question

Working on the fundamental theorem of algebra.

Answer 1

(HINT: A graph of the function h(x) is a good guide to the solution.)

Look for rational zeroes first, that is, factors of ±8, which are 1, 2, 4, 8. Start with x=1:

66-60+137-130-101+80+8=0, so 1 is a zero, so we can divide by it using synthetic division:

1 | 66 -60 137 -130 -101  80    8

     66  66     6  143    13 -88 | -8

     66    6 143    13   -88   -8 |  0 = 66x⁵+6x⁴+143x³+13x²-88x-8.

Note that the magnitude of the constant is still 8. We also have to consider the leading coefficient 66. This has factors: 1, 2, 3, 6, 11, 22, 33, 66. And we still have the remaining factors of 8: 2, 4, 8 (1 has been eliminated already and will not work again, neither will -1). Note that the factors are all even while the factors of 66 contain a mixture of odds and evens. Since the whole expression is a mixture of odds and evens in the coefficients, we can't mix evens in the same factor.

Observe that 66, 88 and 143 are all divisible by 11, so perhaps 1/11 is a zero. Also, 6×11=66 and 13×11=143 and 8×11=88. 6, 13 and 8 are the other coefficients. Let's see if -1/11 is a zero:

-66/11⁵+6/11⁴-143/11³+13/11²+88/11-8=

-6/14641+6/14641-13/121+13/121+8-8=0. Bingo! 11x+1 is a zero.

Divide by 11x+1 using algebraic division:

               6x⁴           13x²        -8

11x+1 ) 66x⁵+6x⁴+143x³+13x² -88x-8

             66x⁵+6x⁴ 143x³+13x² -88x-8

                                      0                    

We are left with 6x⁴+13x²-8=(3x²+8)(2x²-1).

So now we have the means to find the remaining zeroes.

2x²=1 has two zeroes: ±1/√2, that is, -√2/2 and √2/2.

The remaining zeroes are imaginary: ±i√(8/3), or -2i√6/3 and 2i√6/3.

The zeroes are: 1, -1/11, -√2/2, √2/2, -2i√6/3, 2i√6/3.