The problem asks which of the following binomials is a factor of the polynomial $x^3 - 6x^2 + 11x - 6$. The possible binomials are $x-1$, $2x+3$, $x+7$, and $x+1$.

AlgebraPolynomialsFactor TheoremPolynomial FactorizationRoots of Polynomials
2025/3/20

1. Problem Description

The problem asks which of the following binomials is a factor of the polynomial x36x2+11x6x^3 - 6x^2 + 11x - 6. The possible binomials are x1x-1, 2x+32x+3, x+7x+7, and x+1x+1.

2. Solution Steps

We can test each binomial to see if it is a factor using the Factor Theorem. The Factor Theorem states that a polynomial f(x)f(x) has a factor (xa)(x - a) if and only if f(a)=0f(a) = 0.
* Test x1x-1:
If x1x-1 is a factor, then x=1x=1 is a root. We evaluate the polynomial at x=1x=1:
f(1)=(1)36(1)2+11(1)6=16+116=1212=0f(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 12 - 12 = 0.
Since f(1)=0f(1) = 0, x1x-1 is a factor.
* Test 2x+32x+3:
If 2x+32x+3 is a factor, then 2x=32x = -3, so x=32x = -\frac{3}{2} is a root.
f(32)=(32)36(32)2+11(32)6=2786(94)3326=2785443326=27810881328488=31580f(-\frac{3}{2}) = (-\frac{3}{2})^3 - 6(-\frac{3}{2})^2 + 11(-\frac{3}{2}) - 6 = -\frac{27}{8} - 6(\frac{9}{4}) - \frac{33}{2} - 6 = -\frac{27}{8} - \frac{54}{4} - \frac{33}{2} - 6 = -\frac{27}{8} - \frac{108}{8} - \frac{132}{8} - \frac{48}{8} = -\frac{315}{8} \neq 0.
Since f(32)0f(-\frac{3}{2}) \neq 0, 2x+32x+3 is not a factor.
* Test x+7x+7:
If x+7x+7 is a factor, then x=7x = -7 is a root.
f(7)=(7)36(7)2+11(7)6=3436(49)776=343294776=7200f(-7) = (-7)^3 - 6(-7)^2 + 11(-7) - 6 = -343 - 6(49) - 77 - 6 = -343 - 294 - 77 - 6 = -720 \neq 0.
Since f(7)0f(-7) \neq 0, x+7x+7 is not a factor.
* Test x+1x+1:
If x+1x+1 is a factor, then x=1x=-1 is a root.
f(1)=(1)36(1)2+11(1)6=16116=240f(-1) = (-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 = -24 \neq 0.
Since f(1)0f(-1) \neq 0, x+1x+1 is not a factor.
Therefore, x1x-1 is the only binomial in the list that is a factor of the given polynomial.

3. Final Answer

x1x-1

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