Understanding the Equation: Solving y(y² – 5y + 6) = 0 and Why It Equals y(y – 2)(y – 3) = 0

When solving quadratic and polynomial equations, factoring plays a crucial role in simplifying expressions and finding solutions efficiently. One classic example is the equation:
y(y² – 5y + 6) = 0

But why is this equation rewritten as y(y – 2)(y – 3) = 0? And how does factoring help in finding the roots? In this SEO-optimized article, we’ll explore step-by-step how this transformation works and how factoring enables us to solve the equation quickly using the zero-product property.

Understanding the Context

What Does y(y² – 5y + 6) = 0 Mean?

The original equation
y(y² – 5y + 6) = 0
is a product of two factors:

  • Main factor: y
  • Quadratic factor: y² – 5y + 6

Since the product equals zero, the zero-product property tells us that at least one of the factors must be zero:
❌ y = 0
✅ y² – 5y + 6 = 0

y(y^2 - 5y + 6) = 0 \Rightarrow y(y - 2)(y - 3) = 0

Key Insights

This leads us to solving the quadratic, but factoring the quadratic trinomial makes the solution elegant and fast.

Factoring y² – 5y + 6 into (y – 2)(y – 3)

To rewrite y² – 5y + 6, we look for two numbers that:

  • Multiply to +6 (the constant term)
  • Add to –5 (the coefficient of the middle term)

The numbers –2 and –3 satisfy these conditions:
– (–2) × –3 = –6? Wait — correction:
Actually, —2 × –3 = +6 ✅ and
–2 + –3 = –5 ✅

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Final Thoughts

Thus,
y² – 5y + 6 = (y – 2)(y – 3)

Substitute back into the original expression:
y(y² – 5y + 6) = y(y – 2)(y – 3) = 0

Now the equation is fully factored and clearly shown.

Why Is Factoring Important in Solving Equations?

Factoring transforms a potentially complex expression into a product of simpler binomials. Once factored, applying the zero-product property becomes straightforward — each factor is set to zero, unlocking the solutions.

For y(y – 2)(y – 3) = 0,
set each factor equal to zero:

  • y = 0
  • y – 2 = 0 → y = 2
  • y – 3 = 0 → y = 3

Thus, the solutions are y = 0, y = 2, and y = 3

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