Find the rational zeros of the equation 2x4 - x3 - 21x2 plus 9x plus 27 equals 0?

Answer 1

2x4 - x3 - 21x2 + 9x + 27 = 0

Let f(x) = 2x4 - x3 - 21x2 + 9x + 27.

All possible rational zeros = (factors of the constant term, 27)/(factors of the leading coefficient, 2) = ±1, ±3, ±9, ±27)/(±1, ±2) = ±1, ±3, ±9, ±27, ±1/2, ±3/2, ±9/2, ±27/2.

Using synthetic division, test all possible rational zeros:

-1] 2 -1 -21 9 27

(2)(-1) = -2; -1 + -2 = -3

(-3)(-1) = 3; -21 + 3 = -18

(-18)(-1) = 18; 9 + 18 = 27

(27)(-1) = -27; 27 + -27 = 0 (the zero remainder shows that -1 is a zero)

-3] 2 -1 -21 9 27

(2)(-3) = -6; -1 + -6 = -7

(-7)(-3) = 21; -21 + 21 = 0

(0)(-3) = 0; 9 + 0 = 9

(9)(-3) = -27; 27 + -27 = 0 (the zero remainder shows that -3 is a zero)

3/2] 2 -1 -21 9 27

(2)(3/2) = 3; -1 + 3 = 2

(2)(3/2) = 3; -21 + 3 = -18

(-18)(3/2) = -27; 9 + -27 = -18

(-18)(3/2) = -27; 27 + -27 = 0 (the zero remainder shows that 3/2 is a zero)

3] 2 -1 -21 9 27

(2)(3) = 6; -1 + 6 = 5

(5)(3) = 15; -21 + 15 = -6

(-6)(3) = -18; 9 + -18 = -9

(-9)(3) = -27; 27 + -27 = 0 (the zero remainder shows that 3 is a zero)

Thus, the rational zeros are -3, -1, 3/2, and 3.