Question

Determine what number should be added to complete the square of each expression. Then factor each expression. $$ x^{2}+\frac{1}{3} x $$

Short answer

Add \( \frac{1}{36} \); factor as \( \bigg( x + \frac{1}{6} \bigg)^2 - \frac{1}{36} \.

Step-by-step solution

Identify the Coefficient of x

In the expression, identify the coefficient of the linear term, which is \( \frac{1}{3} \).

Divide the Coefficient by 2

Divide the coefficient of \( x \), which is \( \frac{1}{3} \), by 2. This gives \( \frac{1}{3} \div 2 = \frac{1}{6} \).

Square the Result

Square the result from the previous step: \( \bigg( \frac{1}{6} \bigg)^2 = \frac{1}{36} \). This is the number that should be added to complete the square.

Rewrite the Expression

Rewrite the original expression by adding and subtracting \( \frac{1}{36} \): \( x^2 + \frac{1}{3} x + \frac{1}{36} - \frac{1}{36} \).

Factor the Perfect Square Trinomial

The expression \( x^2 + \frac{1}{3} x + \frac{1}{36} \) can be factored as \( \bigg( x + \frac{1}{6} \bigg)^2 \).

Combine and Simplify

After factoring, include the \( - \frac{1}{36} \) outside the perfect square trinomial: \( \bigg( x + \frac{1}{6} \bigg)^2 - \frac{1}{36} \).

Key concepts

Factoring Expressions

Factoring expressions is a foundational skill in algebra. It involves breaking down complex mathematical expressions into simpler parts that can be multiplied to get the original expression. This process is essential because it simplifies equations and makes it easier to solve for unknowns.

To factor an expression like our example, you need to look for common factors. In polynomial expressions, this might mean finding common terms in each element or identifying patterns that allow the expression to be written as a product of simpler polynomials.

For example, if we start with the equation: \[x^2 + \frac{1}{3}x\] and move towards factoring it, we're breaking it into simpler pieces. Each step, like identifying coefficients and perfect squares, helps us factor the expression completely. Remember:

• Identify common factors first
• Apply special factoring rules (like the difference of squares)
• Simplify the expression step by step

Factoring makes solving equations more manageable and provides insight into the structure of the mathematical expression.

Perfect Square Trinomial

A perfect square trinomial is a polynomial of the form \[a^2 + 2ab + b^2\] or \[a^2 - 2ab + b^2\]. These can be quickly factored into \[ (a + b)^2\] or \[ (a - b)^2\] respectively. Understanding how to recognize and form perfect square trinomials simplifies many algebra problems.

In the given exercise, we convert the quadratic \[x^2 + \frac{1}{3}x\] into such a form. Here's how:

• Identify the coefficient of the linear term, which is \ \frac{1}{3} \.
• Divide it by 2 to find \ \frac{1}{6} \.
• Square \[ \frac{1}{6} \] to get \ \frac{1}{36} \.
• Add and subtract this squared value: \ x^2 + \frac{1}{3}x + \frac{1}{36} - \frac{1}{36} \ , making \ x^2 + \frac{1}{3}x + \frac{1}{36} \ a perfect square trinomial

By rewriting and factoring, we see the simpler form: \[ x + \frac{1}{6} \ which is \ \big( x + \frac{1}{6} \big)^2\]. Recognizing perfect square trinomials and knowing how to factor them is a powerful algebra skill.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, generally written as \[ax^2 + bx + c = 0\]. Solving these equations involves several methods: factoring, completing the square, or using the quadratic formula.

In this exercise, we used 'completing the square'. Here’s a recap:

• Start with the equation: \[x^2 + \frac{1}{3}x\]
• Identify 'b' as \ \frac{1}{3} \ and divide by 2: \ \frac{1}{6} \
• Square it: \ \big( \frac{1}{6} \big)^2 = \frac{1}{36} \
• Transform the quadratic into a perfect square trinomial
• Factor the trinomial: \ \big( x + \frac{1}{6} \big)^2 - \frac{1}{36} \

Quadratics can describe various phenomena like projectile motion or area problems. 

Competing the square not only helps in solving but also in understanding the properties of the quadratic graph (a parabola). Each method in solving quadratics gives unique insights into the equation’s behavior and solutions.