Use the remainder theorem to completely factor p(x) = x3 - 6x2 + 11x - 6.
Solution:
Given, f(x) = x3 - 6x2 + 11x - 6
By Rational roots theorem, any rational zeros of P(x) are expressible in the form p/q for integers p,q with p a divisor of the constant term -6 and q a divisor of the coefficient 1 of the leading term.
That means the only possible rational zeros of P(x) are ±1, ±2, ±3, ±4.
Remainder theorem states that the remainder of the division of a polynomial f(x) by a linear polynomial (x - r) is a divisor of f(x) if and only if f(r) = 0.
f(1) = (1)3 - 6(1)2 + 11(1) - 6
= 1 - 6 + 11 - 6
= 12 - 12 = 0
f(1) = 0
(x - 1) is a factor of x3 - 6x2 + 11x - 6
f(-1) = (-1)3 - 6(-1)2 + 11(-1) - 6
= -1 - 6 - 11 - 6
= -24
f(-1) ≠ 0
(x + 1) is not a factor of x3 - 6x2 + 11x - 6
f(2) = (2)3 - 6(2)2 + 11(2) - 6
= 8 - 24 + 22 - 6
= 30 - 30 = 0
f(2) = 0
(x - 2) is a factor of x3 - 6x2 + 11x - 6
f(3) = (3)3 - 6(3)2 + 11(3) - 6
= 27 - 54 + 33 - 6
= 60 - 60 = 0
f(3) = 0
(x - 3) is a factor of x3 - 6x2 + 11x - 6
Therefore, the factors are (x - 1) (x - 2) and (x - 3).
Use the remainder theorem to completely factor p(x) = x3 - 6x2 + 11x - 6.
Summary:
Using the remainder theorem to completely factor p(x) = x3 - 6x2 + 11x - 6. The factors are (x - 1) (x - 2) and (x - 3).
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