y(y^2 - 5y + 6) = 0 \Rightarrow y(y - 2)(y - 3) = 0 - AIKO, infinite ways to autonomy.
Understanding the Equation: Solving y(y² – 5y + 6) = 0 and Why It Equals 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.
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
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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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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
