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# Use the remainder theorem to completely factor p(x) = x^{3} - 6x^{2} + 11x - 6.

**Solution:**

Given, f(x) = x^{3} - 6x^{2} + 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 x^{3 }- 6x^{2} + 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 x^{3} - 6x^{2} + 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 x^{3} - 6x^{2} + 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 x^{3} - 6x^{2} + 11x - 6

Therefore, the factors are (x - 1) (x - 2) and (x - 3).

## Use the remainder theorem to completely factor p(x) = x^{3} - 6x^{2} + 11x - 6.

**Summary:**

Using the remainder theorem to completely factor p(x) = x^{3} - 6x^{2} + 11x - 6. The factors are (x - 1) (x - 2) and (x - 3).

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