(-2, 0) (0, -2) (2, 0) How would I solve this cubic equation without a calculator?

To solve the cubic equation that corresponds to the points \((-2, 0)\), \((0, -2)\), and \((2, 0)\), we need to construct the equation step by step. These points suggest the roots of the equation and give us enough information to find the cubic polynomial.

### Step 1: Identify the roots
Since \((-2, 0)\) and \((2, 0)\) are points where \(y = 0\), these represent the roots of the equation. Therefore, \(x = -2\) and \(x = 2\) are the roots of the equation.

Thus, the equation has factors:
\[
(x + 2)(x - 2)
\]

### Step 2: Construct the polynomial
The cubic equation will have one more root because it's a cubic equation, but from the given points, we can deduce that it also passes through the point \((0, -2)\), which means we don't have a root at \(x = 0\). Instead, this point will help us determine the coefficient of the cubic term.

Using the factors of the equation based on the roots \(x = -2\) and \(x = 2\), the equation has at least these factors:
\[
y = a(x + 2)(x - 2)(x - b)
\]
where \(b\) is a third root, but since no other root is provided, we assume \(b = 0\), meaning there is no additional root. Hence, the equation simplifies to:
\[
y = a(x + 2)(x - 2)
\]

### Step 3: Use the known point \((0, -2)\)
To find the value of \(a\), use the point \((0, -2)\), which means when \(x = 0\), \(y = -2\).

Substitute \(x = 0\) and \(y = -2\) into the equation:
\[
-2 = a(0 + 2)(0 - 2)
\]
\[
-2 = a(2)(-2)
\]
\[
-2 = -4a
\]
Solving for \(a\):
\[
a = \frac{-2}{-4} = \frac{1}{2}
\]

### Step 4: Final cubic equation
Now that we know \(a = \frac{1}{2}\), the equation becomes:
\[
y = \frac{1}{2}(x + 2)(x - 2)
\]

### Step 5: Expand the equation
Now, let's expand the factors:
\[
y = \frac{1}{2}(x^2 - 4)
\]
Thus, the final cubic equation is:
\[
y = \frac{1}{2}(x^2 - 4) = \frac{1}{2}x^2 - 2
\]

So, the cubic equation that passes through the points \((-2, 0)\), \((0, -2)\), and \((2, 0)\) is:
\[
y = \frac{1}{2}x^2 - 2
\]

### Do you have any questions or would you like more details on any step?

Here are 5 related questions you might explore:
1. How do we determine the roots of a cubic equation in general?
2. What if the equation had another distinct root? How would that affect the process?
3. How can we use synthetic division to solve cubic equations?
4. What role do symmetry and intercepts play in solving polynomial equations?
5. How can we use the Factor Theorem for higher-degree polynomials?

**Tip:** Always check your solution by substituting known points back into the equation to verify its accuracy.

Discover how to solve a cubic equation passing through the points (-2, 0), (0, -2), and (2, 0) without a calculator. This step-by-step guide covers algebraic concepts such as polynomial roots, the factor theorem, and equation expansion. Suitable for high school students in Grades 10-12.

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