y(y^2 - 5y + 6) = 0 \Rightarrow y(y - 2)(y - 3) = 0
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.
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
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 ✅
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
How This Equation Ranks in Search Results
This topic — solving polynomial equations using factoring — is highly relevant to algebra learners, students, and educators. A search for phrases like:
- “solve y(y² – 5y + 6) = 0”
- “factoring y(y² – 5y + 6) = 0”
- “root finding y(y – 2)(y – 3) = 0” routinely appears in educational platforms, study guides, and math resources.
Understanding why factoring simplifies solving not only improves problem-solving speed but also boosts SEO relevance for content targeting learners seeking clear, correct, and concise algebraic explanations.
Summary: Key Takeaways
- Start by recognizing the equation y(y² – 5y + 6) = 0 as a product of factors.
- Factor the quadratic: y² – 5y + 6 = (y – 2)(y – 3).
- Rewrite fully: y(y – 2)(y – 3) = 0.
- Apply the zero-product property: each factor equals zero.
- Solve: y = 0, 2, or 3 — these are all real roots.
- Factoring enables clear, efficient solutions and boosts SEO visibility for algebra-related queries.
Further Reading & Related Keywords
- Solving quadratic equations algebraically
- Quick methods for factoring cubic expressions
- Understanding zero-product property and equations
- Step-by-step quadratic equations for beginners
Factor with confidence — understanding this simple equation empowers deeper exploration into algebra and mathematical reasoning!
