Problem 11 - Algebra 2

Question

Simplify. $$ \left(12 y^{2}+36 y+15\right) \div(6 y+3) $$

Step-by-Step Solution

Answer

The simplified expression is \(2y + 5\).

Step 1: Identify Common Factors in the Numerator

We start by factoring the numerator, which is the expression \(12y^2 + 36y + 15\). We look for a common factor in each term. Notice that each coefficient is divisible by 3. Factoring out the 3, we get: \[12y^2 + 36y + 15 = 3(4y^2 + 12y + 5)\]

Step 2: Factor the Quadratic in the Numerator

Next, we focus on factoring the quadratic \(4y^2 + 12y + 5\). We look for two numbers that multiply to \(4 \times 5 = 20\) and add up to 12. These numbers are 10 and 2. We split the middle term accordingly and factor by grouping: \[4y^2 + 12y + 5 = 4y^2 + 10y + 2y + 5 = 2y(2y + 5) + 1(2y + 5)\]This gives us the factored form: \[(2y + 1)(2y + 5)\]

Step 3: Rewrite the Entire Expression

Substituting the factors back, we write the numerator as: \[(12y^2 + 36y + 15) = 3(2y + 1)(2y + 5)\] Now, our division expression becomes: \[\frac{3(2y + 1)(2y + 5)}{6y + 3}\]

Step 4: Factor the Denominator

Factor the denominator \(6y + 3\). Since both terms share a common factor of 3, we factor it out: \[6y + 3 = 3(2y + 1)\]

Step 5: Simplify the Expression

Now that both the numerator and the denominator are factored, cancel out the common factors in the expression. The expression becomes: \[\frac{3(2y + 1)(2y + 5)}{3(2y + 1)}\] Canceling \(3\) and \(2y + 1\), we are left with: \[2y + 5\]

Step 6: Write the Final Simplified Expression

The final simplified form of the original expression is: \[2y + 5\]

Key Concepts

Factoring Polynomials

Factoring polynomials is a key skill in algebra. This process involves expressing a polynomial as a product of its factors, which are polynomials of lower degrees. To start factoring, identify any common factors that can be divided out from all terms in the polynomial. In our exercise, for the polynomial \(12y^2 + 36y + 15\), we noticed that each term shares a factor of 3. Factoring out this 3 simplifies our work and makes it easier to handle the remaining factors.Once the common factor is removed, what's left is a simpler expression that can often be arranged as a product of quadratic or linear factors. This process is systematic:

• Find a common factor among the terms.
• Factor out this common number or variable.
• If a quadratic remains, further factor it typically into two binomials.

This is crucial not just in simplifying expressions, but also in solving polynomial equations, graphing functions, and finding zeros of functions.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing them to their simplest form. This means removing unnecessary parts and combining like terms.In the division problem \((12y^2 + 36y + 15) \div (6y + 3)\), there are multiple steps you take to simplify:

• Factor both the numerator and the denominator to identify shared factors.
• Cancel out the common factors in the numerator and denominator.

This results in a more manageable expression. Simplification helps in calculations, as it transforms a complicated expression with many terms into a sleek and concise result, here, \((2y + 5)\). Always remember, simplifying expressions maintains the original value or equation but makes them easier to work with.

Quadratic Expressions

Quadratic expressions are polynomials of degree two, generally written in the form \(ax^2 + bx + c\). These expressions can be factored into the product of two binomials, a step often needed for solving quadratic equations.When faced with the quadratic part of our original expression, \(4y^2 + 12y + 5\), we look to factor it. To do this:

• Identify two numbers that multiply to \(a \times c\) (in this case, \(4 \times 5 = 20\)) and add up to \(b\) (here, 12).
• In our example, these numbers are 10 and 2. We used them to split the middle term and factor by grouping, resulting in \((2y + 1)(2y + 5)\).

Factoring quadratic expressions reveals roots or solutions of quadratic equations and is foundational in algebra and calculus. The ability to factor effectively makes solving equations and analyzing functions much easier.