Problem 53 - Elementary and Intermediate Algebra

Question

The following is a list of random factoring problems. Factor each expression. If an expression is not factorable, write "prime." See Examples 1-5. $$ 4 x^{2} y^{2}+4 x y^{2}+y^{2} $$

Step-by-Step Solution

Answer

The expression factors to \( y^2(2x + 1)^2 \).

Step 1: Identify Common Factors

Look at the expression \(4x^2y^2 + 4xy^2 + y^2\). Notice that each term contains a \(y^2\). Factor \(y^2\) out of the entire expression.\[ y^2(4x^2 + 4x + 1) \]

Step 2: Factor the Quadratic Expression

Now focus on the quadratic expression inside the parentheses: \(4x^2 + 4x + 1\). This expression is in the form of a trinomial quadratic equation \(ax^2 + bx + c\).

Step 3: Apply the Quadratic Formula

Since factoring by inspection or other methods might not be straightforward, let's find the roots, if any, by using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 4\), \(b = 4\), and \(c = 1\):\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]\[ x = \frac{-4 \pm \sqrt{16 - 16}}{8} \]\[ x = \frac{-4 \pm 0}{8} \]\[ x = -\frac{1}{2} \]

Step 4: Factor Using the Roots

Since the quadratic has a double root \( x = -\frac{1}{2} \), the expression \(4x^2 + 4x + 1\) can be factored as:\[ (2x + 1)^2 \]

Step 5: Write the Final Factored Expression

Combine the factors obtained in previous steps: \(y^2(2x + 1)^2\). Therefore, the completely factored expression for the original problem is:\[ y^2(2x + 1)^2 \]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). This equation can be factored or solved using the formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This means you substitute the coefficients \(a\), \(b\), and \(c\) from your quadratic equation into the formula.
The result gives you the roots or solutions of the equation.
To illustrate, let's take the quadratic expression \(4x^2 + 4x + 1\) from our example. Here, identify the coefficients as \(a = 4\), \(b = 4\), and \(c = 1\). Plug these into the quadratic formula:

• \(x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}\)
• Simplifying the equation results in: \( x = -\frac{1}{2} \)

The "\(\pm\)" symbol indicates potential multiple roots. However, in this case, both calculations yield the same root \(-\frac{1}{2}\). This is known as a double root.

Trinomial Quadratic Equation

A trinomial quadratic equation is an equation of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. It features three terms: one squared term, one linear term, and a constant term.
The equation from our exercise, \(4x^2 + 4x + 1\), falls under this category.

• *"Trinomial"* implies three terms are being considered.

Trinomial quadratic equations are typically solved by:

• Factoring, if possible;
• Using the quadratic formula when factoring is complex or not straightforward.

In our example, direct factoring might not be easy because the coefficients and constant do not lend themselves to simple division by common factors.
Instead, using the quadratic formula was effective in determining the root \(x = -\frac{1}{2}\). Once you've identified the root through the formula, you can express the trinomial in factored form, which was found in our solution to be \((2x + 1)^2\).

Common Factor Identification

Identifying common factors is often the first step in simplifying or factoring expressions. When factoring expressions, always check if a common factor is present in all terms.
In the example \(4x^2y^2 + 4xy^2 + y^2\), notice that each term contains a factor of \(y^2\). By factoring out \(y^2\), the expression simplifies to:

• \(y^2(4x^2 + 4x + 1)\)

This approach reduces the complexity of the expression, making further factoring easier to analyze. Check for the greatest common factor (GCF) by:

• Looking at coefficients and variable powers.shared among all terms;
• Determining the highest power of each shared variable.

Always simplify expressions as much as possible before tackling more complex operations like applying the quadratic formula or advanced factoring methods.
Identification of common factors not only simplifies the problem but also sets a solid foundation for solving or factoring more sophisticated algebraic expressions.