Answer:
-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 <em>rational root theorem</em>, 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 <em>synthetic division</em>.
7|6 -41 -8 7
|<u> 42 7 -7
</u>
6 1 -1 0
So, x = 7 is a zero, and (6x³ - 41x² - 8x + 7)/(x - 7) = 6x² + x – 1
<em>Solve the quadratic</em>.
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.
