Problem 97 - Algebra and Trigonometry with Analytic Geometry

Question

Factor the polynomial. $$ y^{2}-x^{2}+8 y+16 $$

Step-by-Step Solution

Answer

The factored form is \( (y + x + 4)(y - x + 4) \).

Step 1: Rearrange the polynomial terms

Begin by rearranging the terms of the polynomial into a form that will make it easier to factor. Rearrange the polynomial as \( y^2 + 8y + 16 - x^2 \). This separates the \( y \)-terms and can potentially help in recognizing patterns like perfect squares.

Step 2: Focus on y-terms and identify a perfect square trinomial

Notice the expression \( y^2 + 8y + 16 \) is a trinomial. Check if it can be written as a square of a binomial. Observe that \( y^2 + 8y + 16 \) is in the form \( (y + a)^2 \) where \( a^2 = 16 \) and \( 2ay = 8y \). Here, \( a = 4 \), so \( y^2 + 8y + 16 = (y+4)^2 \).

Step 3: Express the polynomial as a difference of squares

The rearranged polynomial becomes \( (y+4)^2 - x^2 \). Recognize this form as a difference of squares. Recall that a difference of squares can be factored using the identity \( a^2 - b^2 = (a + b)(a - b) \).

Step 4: Apply the difference of squares formula

Let \( a = y+4 \) and \( b = x \) in the difference of squares formula. Therefore, the expression \( (y+4)^2 - x^2 \) can be factored as \( (y+4+x)(y+4-x) \).

Step 5: Write the final factored form

The polynomial \( y^2 - x^2 + 8y + 16 \) is thus factored into \( (y + x + 4)(y - x + 4) \).

Key Concepts

Difference of Squares

The difference of squares is a widespread factoring technique in algebra. Whenever you have a polynomial in the form \( a^2 - b^2 \), you can factor it into \((a + b)(a - b)\). This works because when you multiply \( (a + b)(a - b) \), the middle terms \(ab\) and \(-ab\) cancel out.For example, if you have the expression \( y^2 - x^2 \), you can immediately recognize it as a difference of squares. Here, \( y \) and \( x \) are squared, so \( a = y \) and \( b = x \), and it factors into \((y + x)(y - x)\).
In the exercise, once the trinomial \((y+4)^2\) is formed, it becomes part of a difference of squares together with \( x^2 \). Thus, \((y+4)^2 - x^2\) can be broken down using this simple yet powerful method. Remember, recognizing this pattern makes factoring much quicker!

Perfect Square Trinomial

Recognizing a perfect square trinomial is crucial in factoring processes. It often occurs when you see an expression of the form \( a^2 + 2ab + b^2 \). This always factors into \((a + b)^2\) or \((a - b)^2\).
For instance, if given \( y^2 + 8y + 16 \), check if it fits the pattern. Notice that the first term is \( y^2 \) and the last term is \( 16 \), which is a perfect square, \( 4^2 \). The middle term is \( 8y \), which can be expressed as \( 2 \times y \times 4 \). Thus this trinomial fits into \((y + 4)^2\).
Spotting these quickly means less math struggle! It helps transform complex expressions into something more manageable, eventually simplifying the whole problem.

Factoring Techniques

Factoring techniques are essential tools when dealing with polynomials in algebra. Different polynomials require different methods based on their structure.

• Identify factorable patterns like difference of squares, perfect squares, or the greatest common factor (GCF).
• Look for patterns in the terms, as seen in perfect square trinomials.
• Combine techniques if necessary. For instance, in this exercise, you first identify a perfect square trinomial \((y^2 + 8y + 16)\), then notice it's part of a difference of squares with \(x^2\).

Being flexible with these techniques allows you to tackle any polynomial and break it down into simpler factors. Understanding when and how to use these techniques is a game-changer in algebra, making seemingly complex problems solvable!