Find all rational roots for P(x)=0
P(x) = 6x^4 - 13x^2 - 39x - 15
To find the rational roots of a polynomial, we can use the rational root theorem. According to the theorem, any rational root of a polynomial equation can be written as a quotient of a factor of the constant term divided by a factor of the leading coefficient.
The constant term of P(x) is -15, which has the factors ±1, ±3, ±5, ±15.
The leading coefficient of P(x) is 6, which has the factors ±1, ±2, ±3, ±6.
The possible rational roots are the combinations of these factors:
±1/1, ±3/1, ±5/1, ±15/1, ±1/2, ±3/2, ±5/2, ±15/2, ±1/3, ±3/3, ±5/3, ±15/3, ±1/6, ±3/6, ±5/6, ±15/6
Now we can go through each of these possible rational roots and check if they satisfy P(x) = 0.
When we substitute ±1 into P(x), we get:
P(1) = 6(1)^4 - 13(1)^2 - 39(1) - 15 = -61 ≠ 0
P(-1) = 6(-1)^4 - 13(-1)^2 - 39(-1) - 15 = 7 ≠ 0
When we substitute ±3 into P(x), we get:
P(3) = 6(3)^4 - 13(3)^2 - 39(3) - 15 = 519 ≠ 0
P(-3) = 6(-3)^4 - 13(-3)^2 - 39(-3) - 15 = -1295 ≠ 0
When we substitute ±5 into P(x), we get:
P(5) = 6(5)^4 - 13(5)^2 - 39(5) - 15 = 19975 ≠ 0
P(-5) = 6(-5)^4 - 13(-5)^2 - 39(-5) - 15 = -22705 ≠ 0
When we substitute ±15 into P(x), we get:
P(15) = 6(15)^4 - 13(15)^2 - 39(15) - 15 = 123795 ≠ 0
P(-15) = 6(-15)^4 - 13(-15)^2 - 39(-15) - 15 = -89625 ≠ 0
When we substitute ±1/2 into P(x), we get:
P(1/2) = 6(1/2)^4 - 13(1/2)^2 - 39(1/2) - 15 = 0
Therefore, the rational root of P(x) = 0 is x = 1/2.