y^2 \cdot y - 7y^2 + 10y = 0 \Rightarrow y^3 - 7y^2 + 10y = 0 \Rightarrow y(y^2 - 7y + 10) = 0

February 22, 2026

$y^2 \cdot y - 7y^2 + 10y = 0 \Rightarrow y^3 - 7y^2 + 10y = 0 \Rightarrow y(y^2 - 7y + 10) = 0$

Understanding the Equation: Solving y²·y – 7y² + 10y = 0 and Factorization

When solving polynomial equations, breaking them down step by step is essential for clarity and accuracy. One such equation often discussed in algebra is:

y²·y – 7y² + 10y = 0

At first glance, this may seem daunting, but simplifying and factoring reveals its underlying structure. Let’s explore how this equation transforms and how to solve it efficiently.

Step 1: Simplify the Equation

The expression starts with:

y²·y – 7y² + 10y

Recall that multiplying y² by y gives:

y³ – 7y² + 10y

Thus, the equation simplifies to: y³ – 7y² + 10y = 0

This is a cubic equation that can be solved using factoring techniques.

Step 2: Factor Out the Common Term

Notice that every term contains at least one factor of y. Factoring out y gives:

y(y² – 7y + 10) = 0

This step is crucial because it reduces the problem from solving a cubic to solving a quadratic equation inside parentheses, which is much simpler.

Step 3: Factor the Quadratic Expression

Now consider the quadratic: y² – 7y + 10

We seek two numbers that multiply to 10 and add to –7. These numbers are –5 and –2.

Therefore: y² – 7y + 10 = (y – 5)(y – 2)

Substitute this back: y(y – 5)(y – 2) = 0

Step 4: Solve for the Roots

Set each factor equal to zero:

y = 0
y – 5 = 0 → y = 5
y – 2 = 0 → y = 2

Conclusion

The equation y²·y – 7y² + 10y = 0 simplifies cleanly to y(y² – 7y + 10) = 0, and further factoring leads to the solutions y = 0, y = 2, and y = 5. This type of step-by-step algebraic simplification and factoring is vital in solving higher-degree equations and forming a strong foundation in polynomial analysis.

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Tip: When working with polynomial equations, always look for common factors first—like factoring out y—to reduce complexity and avoid unnecessary steps. Mastering this process improves problem-solving speed and accuracy in algebra.