find all rational roots of the polynomial 6x^3-13x^2-41x-12=0
any body? need help a.s.a.p
To find the rational roots of a polynomial, we can use the Rational Root Theorem. According to this theorem, if a rational number (p/q) is a root of a polynomial, then p must be a factor of the constant term (in this case, -12), and q must be a factor of the leading coefficient (in this case, 6).
The factors of the constant term -12 are ±1, ±2, ±3, ±4, ±6, and ±12.
The factors of the leading coefficient 6 are ±1, ±2, ±3, and ±6.
Using these factors, we can test each possible rational root by substituting it into the polynomial and checking if it equals zero. If we find any roots, they will be rational.
Let's evaluate the polynomial using each of these rational numbers:
1/1: 6(1/1)^3 - 13(1/1)^2 - 41(1/1) - 12 = 6 - 13 - 41 - 12 = -60 (not zero)
1/2: 6(1/2)^3 - 13(1/2)^2 - 41(1/2) - 12 = 3/8 - 13/4 - 41/2 - 12 = -454/8 (not zero)
1/3: 6(1/3)^3 - 13(1/3)^2 - 41(1/3) - 12 = 2/27 - 13/9 - 41/3 - 12 = -503/27 (not zero)
1/6: 6(1/6)^3 - 13(1/6)^2 - 41(1/6) - 12 = 1/216 - 13/36 - 41/6 - 12 = -927/216 (not zero)
2/1: 6(2/1)^3 - 13(2/1)^2 - 41(2/1) - 12 = 48 - 52 - 82 - 12 = -98 (not zero)
2/2: 6(2/2)^3 - 13(2/2)^2 - 41(2/2) - 12 = 6 - 13 - 41 - 12 = -60 (not zero)
2/3: 6(2/3)^3 - 13(2/3)^2 - 41(2/3) - 12 = 16/27- 52/9 - 82/3 - 12 = -98/27 (not zero)
2/6: 6(2/6)^3 - 13(2/6)^2 - 41(2/6) - 12 = 1/27 - 13/9 - 41/3 - 12 = -185/54 (not zero)
...
Continuing the process with the remaining factors, you will find that none of these values make the polynomial equal to zero.
Therefore, the polynomial 6x^3 - 13x^2 - 41x - 12 = 0 does not have any rational roots.