Find all rational zeros of the function h(x)=6x^3-41x^2-8x+7

-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.

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )

26?

step-by-step explanation: