Find all rational zeros of the function

h (x) = 6x^3-41x^2-8x+7

X = 7, 1/3,-1/2

The zero are those above

-½, ⅓, 7

Step-by-step explanation:

The general formula for a third-degree polynomial is

f (x) = ax³ + bx² + cx + d

Your polynomial is

h (x) = 6x³ - 41x² - 8x + 7

a = 6; d = 7

According to the rational root theorem, the possible rational roots are

Factors of d/Factors of a

Factors of d = ±1, ±7

Factors of a = ±1, ±2, ±3, ±6

Potential roots are x = ±1/1, ±1/2, ±1/3, ±1/6, ±7/1, ±7/2, ±7/3, ±7/6

Putting them in order, we get the potential roots

x = - 7, - ⁷/₂, - ⁷/₃, - ⁷/₆, - 1, - ½, - ⅓, - ⅙, ⅙, ⅓, ½, 1, ⁷/₆, ⁷/₃, ⁷/₂, 7

Now, it's a matter of trial and error to find a zero.

Let's try x = 7 by synthetic division.

7|6 - 41 - 8 7

| 42 7 - 7

6 1 - 1 0

So, x = 7 is a zero, and (6x³ - 41x² - 8x + 7) / (x - 7) = 6x² + x - 1

Solve the quadratic.

6x² + x - 1 = 0

(3x - 1) (2x + 1) = 0

3x - 1 = 0 2x + 1 = 0

3x = 1 2x = - 1

x = ⅓ x = - ½

The zeroes of h (x) are - ½, ⅓, and 7.