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

Understanding the Context

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:

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

Key Insights

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.

Continue Reading

A plant biologist models root growth with the function $ r(t) = \sqrt{t} + 2 $. What is the root growth at $ t = 9 $ weeks?
Substitute $ t = 9 $: $ r(9) = \sqrt{9} + 2 = 3 + 2 = 5 $.
A science administrator tracks grant applications over 5 years: 120, 135, 150, 165, and 180. What is the average number of applications per year?

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Final Thoughts

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)