Find any rational roots of P(x).
45. P(x) = x^3 + 5x^2 + x + 5
To find the rational roots of P(x) = x^3 + 5x^2 + x + 5, we can use the Rational Root Theorem. According to the theorem, any rational root of the polynomial has the form of p/q, where p is a factor of the constant term (5) and q is a factor of the leading coefficient (1).
The factors of 5 are ±1 and ±5, and the factors of 1 are ±1.
Therefore, the possible rational roots of P(x) are: ±1/1, ±5/1, ±1/5, and ±5/5.
Simplifying these roots, we have: ±1, ±5, ±1/5.
Now, we can substitute these possible rational roots into P(x) and see if any of them make the polynomial equal to zero.
For P(x) = x^3 + 5x^2 + x + 5:
Substituting x = 1 gives: P(1) = 1^3 + 5(1)^2 + 1 + 5 = 1 + 5 + 1 + 5 = 12
Substituting x = -1 gives: P(-1) = (-1)^3 + 5(-1)^2 - 1 + 5 = -1 + 5 - 1 + 5 = 8
Substituting x = 5 gives: P(5) = 5^3 + 5(5)^2 + 5 + 5 = 125 + 125 + 5 + 5 = 260
Substituting x = -5 gives: P(-5) = (-5)^3 + 5(-5)^2 - 5 + 5 = -125 + 125 - 5 + 5 = 0
Substituting x = 1/5 gives: P(1/5) = (1/5)^3 + 5(1/5)^2 + 1/5 + 5 = 1/125 + 1/5 + 1/5 + 5 = 137/125
Substituting x = -1/5 gives: P(-1/5) = (-1/5)^3 + 5(-1/5)^2 - 1/5 + 5 = -1/125 + 1/5 - 1/5 + 5 = 524/125
From the substitutions, we can see that the only rational root is x = -5.