Factoring: Reverse Engineering
You’ve learned how to expand expressions — turning (x + 2)(x + 3) into x^2 + 5x + 6. Factoring is the reverse: starting with x^2 + 5x + 6 and breaking it back into (x + 2)(x + 3). But why would you want to do that?
Because factored form reveals the roots.
The Big Idea
When a quadratic is written in factored form:
You can instantly read off where the parabola crosses the x-axis. Those crossing points are the roots (also called zeros): x = r1 and x = r2.
Why? Because if x = r1, then the first factor becomes (r1 - r1) = 0, and anything times zero is zero. So y = 0. Same logic for x = r2.
Part 1: Factored Form — See the Roots Directly
Let’s start with the factored form and see how the roots control the parabola:
Drag the sliders and watch:
- The parabola always crosses the x-axis at exactly r1 and r2
- Move the roots farther apart and the parabola gets wider
- Move them closer together and it gets narrower
- What happens when r1 = r2? The parabola just touches the x-axis at one point — a double root!
Part 2: From Factored to Standard Form
When you expand (x - r1)(x - r2), you get the standard form. Let’s watch the expansion happen live:
The factored form:
Expands to the standard form (using FOIL):
The pattern is always the same:
- The coefficient of x equals -(r1 + r2) — the negative sum of the roots
- The constant term equals r1 times r2 — the product of the roots
This is why factoring works! If you can find two numbers that ADD to give the middle coefficient and MULTIPLY to give the constant, those are your roots.
Let’s verify both forms produce the same graph:
Part 3: The Connection Between Roots and Vertex
The vertex (the lowest or highest point of the parabola) always sits exactly halfway between the two roots:
Set r1 = -3 and r2 = 1. The roots are at x = -3 and x = 1. Halfway between them is x = (-3 + 1)/2 = -1. Look at the vertex — it’s right there at x = -1!
This always works because parabolas are symmetric. The axis of symmetry passes through the vertex and splits the two roots evenly.
Part 4: Adding a Leading Coefficient
What if there’s a number in front? y = a(x - r1)(x - r2) still has the same roots, but the a controls how steep and which direction:
Experiment with a:
- a > 0: Parabola opens upward (U-shape)
- a < 0: Parabola opens downward (upside-down U)
- |a| > 1: Narrower and steeper
- |a| < 1: Wider and flatter
- The roots DON’T change! The value of a stretches the curve but keeps the same x-crossings.
Part 5: The Factoring Process
So far we’ve gone from factored form to graph. In practice, you often go the other direction: starting with standard form and trying to factor.
For x^2 + bx + c, you need two numbers r1 and r2 such that:
- r1 + r2 = -b (they add to the negative of the middle coefficient)
- r1 * r2 = c (they multiply to the constant term)
Example: Factor x^2 - 5x + 6.
You need two numbers that multiply to 6 and add to 5. Those are 2 and 3. Since the middle term is -5x, the roots are positive: (x - 2)(x - 3).
Let’s verify:
The roots are at x = 2 and x = 3. Factoring confirmed!
Practice factoring these:
- x^2 + 5x + 6 (find two numbers that add to 5 and multiply to 6)
- x^2 - x - 12 (add to -1, multiply to -12)
- x^2 - 9 (this is a special case — difference of squares!)
Answers: (x + 2)(x + 3), (x - 4)(x + 3), (x - 3)(x + 3)
Part 6: Special Factoring Patterns
Some quadratics factor in recognizable patterns:
Difference of squares:
The roots are always symmetric: one at +a and one at -a. The vertex is always at the origin’s x-value (x = 0).
Perfect square trinomial:
This is the double root case — the parabola just touches the x-axis at one point.
Wrapping Up
| Concept | Key Takeaway |
|---|---|
| Factored form | y = (x - r1)(x - r2) shows roots directly |
| Roots | Where the graph crosses the x-axis (y = 0) |
| Sum of roots | Equals -b/a (negative of middle coefficient over leading) |
| Product of roots | Equals c/a (constant over leading coefficient) |
| Vertex | Halfway between the two roots |
| Leading coefficient (a) | Controls steepness and direction, not roots |
Final Challenge: A parabola has its vertex at (1, -4) and passes through (3, 0). Find the other root, write the factored form, and expand to standard form.
Hint: If the vertex is at x = 1 and one root is at x = 3, the other root is the same distance on the other side of the vertex. So it’s at x = -1. Factored form: y = (x - 3)(x + 1) = x^2 - 2x - 3.
Factoring is like reverse engineering. When you see a polynomial and can break it into factors, you’ve cracked its code — you know exactly where it hits zero, where its vertex is, and what its graph looks like. That’s a superpower.