Solve Algebraic Inequalities: A Step-by-Step Guide

Hey algebra whizzes! Today, we're diving deep into the nitty-gritty of solving inequalities. Specifically, we're tackling a doozy that looks a bit intimidating at first glance: $\frac{(2x + x^2 + 3)(x + 2)}{x^2 + 3x + 2} \le -1 - \frac{x+3}{x-2}$. Now, I know what you might be thinking, "Whoa, that's a lot of terms!" But trust me, guys, once we break it down, it's totally manageable. We're going to go through this step-by-step, making sure you understand every little bit of it. So, grab your notebooks, get comfy, and let's conquer this inequality together!

Step 1: Simplify and Factor

The first move in any complex algebraic problem, especially when dealing with inequalities, is to simplify. This means factoring everything we possibly can. Let's look at the left side of our inequality: $\frac{(2x + x^2 + 3)(x + 2)}{x^2 + 3x + 2}$. The numerator has a quadratic expression, $x^2 + 2x + 3$. Let's see if this quadratic can be factored. We're looking for two numbers that multiply to 3 and add to 2. Hmm, doesn't look like it factors easily with integers. We can check the discriminant ($b^2 - 4ac$) to see if it has real roots. Here, $a=1$, $b=2$, $c=3$, so the discriminant is $2^2 - 4(1)(3) = 4 - 12 = -8$. Since the discriminant is negative, $x^2 + 2x + 3$ has no real roots and is always positive (because the coefficient of $x^2$ is positive). This is a super important piece of information, guys! It means this part of the expression will never make the inequality flip or cause division by zero issues related to its own roots.

Now, let's look at the denominator of the left side: $x^2 + 3x + 2$. This one is much friendlier! We need two numbers that multiply to 2 and add to 3. Bingo! That's 1 and 2. So, $x^2 + 3x + 2 = (x+1)(x+2)$.

So, the left side becomes: $\frac{(x^2 + 2x + 3)(x + 2)}{(x+1)(x+2)}$.

Notice that we have an $(x+2)$ term in both the numerator and the denominator. Awesome! We can cancel these out, but we have to be careful. We can only cancel if $x+2 \ne 0$, which means $x \ne -2$. We'll need to remember this restriction.

After canceling, the left side simplifies to: $\frac{x^2 + 2x + 3}{x+1}$, with the condition that $x \ne -2$.

Now let's tackle the right side of the inequality: $-1 - \frac{x+3}{x-2}$. We can combine these terms by finding a common denominator, which is $(x-2)$.

So, $-1 - \frac{x+3}{x-2} = -\frac{x-2}{x-2} - \frac{x+3}{x-2} = \frac{-(x-2) - (x+3)}{x-2} = \frac{-x + 2 - x - 3}{x-2} = \frac{-2x - 1}{x-2}$.

Our inequality has now transformed into: $\frac{x^2 + 2x + 3}{x+1} \le \frac{-2x - 1}{x-2}$, with the restriction that $x \ne -2$. We also need to remember the inherent restrictions from the denominators: $x+1 \ne 0$ (so $x \ne -1$) and $x-2 \ne 0$ (so $x \ne 2$).

Step 2: Move All Terms to One Side

The golden rule when solving inequalities (and equations, for that matter) is to get everything on one side. This way, we're comparing an expression to zero, which makes analysis much easier. So, let's move the right side to the left side:

$\frac{x^2 + 2x + 3}{x+1} - \frac{-2x - 1}{x-2} \le 0$

Combine the fractions by finding a common denominator, which is $(x+1)(x-2)$:

$\frac{(x^2 + 2x + 3)(x-2)}{(x+1)(x-2)} - \frac{(-2x - 1)(x+1)}{(x+1)(x-2)} \le 0$

Now, let's expand the numerators:

First numerator: $(x^2 + 2x + 3)(x-2) = x^3 - 2x^2 + 2x^2 - 4x + 3x - 6 = x^3 - x - 6$

Second numerator: $(-2x - 1)(x+1) = -2x^2 - 2x - x - 1 = -2x^2 - 3x - 1$

Now, subtract the second expanded numerator from the first:

$(x^3 - x - 6) - (-2x^2 - 3x - 1) = x^3 - x - 6 + 2x^2 + 3x + 1 = x^3 + 2x^2 + 2x - 5$

So, our inequality is now:

$\frac{x^3 + 2x^2 + 2x - 5}{(x+1)(x-2)} \le 0$

We also need to remember our restrictions: $x \ne -2$, $x \ne -1$, and $x \ne 2$.

Step 3: Analyze the Numerator and Denominator

Our next mission is to find the roots of the numerator, $P(x) = x^3 + 2x^2 + 2x - 5$, and the roots of the denominator, $Q(x) = (x+1)(x-2)$. The roots of the denominator are clearly $x = -1$ and $x = 2$. These are the points where our expression is undefined.

Finding the roots of a cubic polynomial like $x^3 + 2x^2 + 2x - 5$ can be tricky. We usually try to find rational roots using the Rational Root Theorem. This theorem states that any rational root $p/q$ must have $p$ as a factor of the constant term (-5) and $q$ as a factor of the leading coefficient (1). So, possible rational roots are $\pm 1, \pm 5$.

Let's test these values:

• For $x=1$: $1^3 + 2(1)^2 + 2(1) - 5 = 1 + 2 + 2 - 5 = 0$. Bingo! $x=1$ is a root.

Since $x=1$ is a root, $(x-1)$ must be a factor of $x^3 + 2x^2 + 2x - 5$. We can use polynomial division or synthetic division to find the other factor.

Using synthetic division with root 1:

1 | 1   2   2   -5
  |     1   3    5
  ----------------
    1   3   5    0

The quotient is $x^2 + 3x + 5$. So, the numerator factors as $(x-1)(x^2 + 3x + 5)$.

Now, let's analyze the quadratic factor $x^2 + 3x + 5$. We check its discriminant: $b^2 - 4ac = 3^2 - 4(1)(5) = 9 - 20 = -11$. Since the discriminant is negative and the leading coefficient (1) is positive, $x^2 + 3x + 5$ is always positive for all real values of $x$. This is fantastic news, guys! It means this quadratic factor never affects the sign of our inequality.

So, our inequality $\frac{x^3 + 2x^2 + 2x - 5}{(x+1)(x-2)} \le 0$ simplifies to analyzing:

$\frac{(x-1)(x^2 + 3x + 5)}{(x+1)(x-2)} \le 0$

Since $(x^2 + 3x + 5)$ is always positive, we only need to consider the sign of $\frac{x-1}{(x+1)(x-2)}$.

Step 4: Use a Sign Table (or Number Line)

Now we have a much simpler expression to analyze: $\frac{x-1}{(x+1)(x-2)} \le 0$. The critical points are the values of $x$ where the numerator or denominator equals zero. These are $x=1$ (from $x-1=0$), $x=-1$ (from $x+1=0$), and $x=2$ (from $x-2=0$). Remember our extra restriction $x \ne -2$ from the original simplification.

These critical points divide the number line into intervals: $(-\infty, -2)$, $(-2, -1)$, $(-1, 1]$, $[1, 2)$, and $(2, \infty)$. Note that $x=1$ can be included because the inequality is 'less than or equal to', but $x=-1$ and $x=2$ cannot be included because they make the denominator zero.

Let's create a sign table to check the sign of $\frac{x-1}{(x+1)(x-2)}$ in each interval:

Step 5: State the Solution

Based on our sign table, the expression $\frac{x-1}{(x+1)(x-2)}$ is less than or equal to zero in the intervals $(-\infty, -2)$, $(-2, -1)$, and $[1, 2)$.

Combining these intervals, the solution set is $(-\infty, -2) \cup (-2, -1) \cup [1, 2)$.

And there you have it, guys! We've successfully navigated through a complex inequality. Remember, the key is to stay organized, simplify as much as possible, factor carefully, and use a sign table to analyze the intervals. Don't be intimidated by the looks of a problem; break it down into smaller, manageable steps, and you'll be solving them like a pro. Keep practicing, and you'll master these in no time! Happy solving!