Question

abc>0 and a+b+c<0 and |a|/a + |b|/b + |c|/c =x find x³-6x²+11x-6

Answer 1

-24

Answer 2

The given conditions are:

1. $abc > 0$
2. $a+b+c < 0$
3. $x = \frac{|a|}{a} + \frac{|b|}{b} + \frac{|c|}{c}$

The term $\frac{|y|}{y}$ is equal to 1 if $y > 0$ and -1 if $y < 0$ (for $y \neq 0$).
So, $x$ is the sum of three terms, each being either 1 or -1.

Let's analyze the signs of a, b, and c based on the given conditions.
Condition 1, $abc > 0$, implies either all three numbers are positive, or one number is positive and the other two are negative.

Case 1: $a > 0$, $b > 0$, $c > 0$.
In this case, $a+b+c > 0$. This contradicts condition 2 ($a+b+c < 0$). So, this case is not possible.

Case 2: One of the numbers is positive, and the other two are negative.
Let's consider the possibilities:

• Possibility 2a: $a > 0$, $b < 0$, $c < 0$.
  $abc = (+)(-) (-) = + > 0$. This satisfies condition 1.
  $a+b+c$ can be negative. For example, if $a=1, b=-2, c=-2$, then $a+b+c = 1-2-2 = -3 < 0$. This satisfies condition 2.
  This combination of signs is possible.

• Possibility 2b: $a < 0$, $b > 0$, $c < 0$.
  $abc = (-) (+) (-) = + > 0$. This satisfies condition 1.
  $a+b+c$ can be negative. For example, if $a=-2, b=1, c=-2$, then $a+b+c = -2+1-2 = -3 < 0$. This satisfies condition 2.
  This combination of signs is possible.

• Possibility 2c: $a < 0$, $b < 0$, $c > 0$.
  $abc = (-) (-) (+) = + > 0$. This satisfies condition 1.
  $a+b+c$ can be negative. For example, if $a=-2, b=-2, c=1$, then $a+b+c = -2-2+1 = -3 < 0$. This satisfies condition 2.
  This combination of signs is possible.

The only scenario consistent with both conditions $abc > 0$ and $a+b+c < 0$ is that exactly one of the numbers a, b, c is positive and the other two are negative.

Now let's find the value of x in this scenario.
$x = \frac{|a|}{a} + \frac{|b|}{b} + \frac{|c|}{c}$.
If a number is positive, its corresponding term is 1.
If a number is negative, its corresponding term is -1.
In the scenario where one number is positive and two are negative, the terms in the sum for x will be one 1 and two -1s.
For example, if $a > 0, b < 0, c < 0$, then $\frac{|a|}{a}=1, \frac{|b|}{b}=-1, \frac{|c|}{c}=-1$.
$x = 1 + (-1) + (-1) = 1 - 1 - 1 = -1$.
If $a < 0, b > 0, c < 0$, then $\frac{|a|}{a}=-1, \frac{|b|}{b}=1, \frac{|c|}{c}=-1$.
$x = -1 + 1 + (-1) = -1$.
If $a < 0, b < 0, c > 0$, then $\frac{|a|}{a}=-1, \frac{|b|}{b}=-1, \frac{|c|}{c}=1$.
$x = -1 + (-1) + 1 = -1$.

In all valid cases, the value of x is -1.

Now we need to find the value of the expression $x^3 - 6x^2 + 11x - 6$ when $x = -1$.
Substitute $x = -1$ into the expression:
$(-1)^3 - 6(-1)^2 + 11(-1) - 6$
$= -1 - 6(1) - 11 - 6$
$= -1 - 6 - 11 - 6$
$= -7 - 11 - 6$
$= -18 - 6$
$= -24$.

Alternatively, we can factor the polynomial $P(x) = x^3 - 6x^2 + 11x - 6$. By testing small integer values, we find that $P(1) = 1-6+11-6=0$, $P(2) = 8-24+22-6=0$, and $P(3) = 27-54+33-6=0$.
So the polynomial can be factored as $(x-1)(x-2)(x-3)$.
Evaluating this at $x=-1$:
$(-1-1)(-1-2)(-1-3) = (-2)(-3)(-4) = 6(-4) = -24$.

The final answer is $\boxed{-24}$.