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Chapter 1: Problem 39

In \(27-39,\) factor each polynomial completely. $$ x^{2} y-16 y $$

Short Answer

Expert verified
The polynomial \(x^2y - 16y\) factors completely to \(y(x + 4)(x - 4)\).

Step by step solution

01

Identify Common Factors

Examine each term in the polynomial \(x^2y - 16y\). We notice that \(y\) is a common factor in both terms because it appears in both \(x^2y\) and \(-16y\).
02

Factor Out the Common Factor

Factor \(y\) out of the polynomial. Write:\[x^2y - 16y = y(x^2 - 16)\]
03

Recognize and Apply the Difference of Squares Formula

The expression \(x^2 - 16\) is a difference of squares. Recall the formula for difference of squares: \(a^2 - b^2 = (a + b)(a - b)\). Here, \(x^2\) is \(a^2\) and \(16\) is \(4^2\). Apply the difference of squares:\[x^2 - 16 = (x + 4)(x - 4)\]
04

Write the Completely Factored Form

Combine what was factored in Step 2 with the factoring done in Step 3. The fully factored polynomial is:\[y(x + 4)(x - 4)\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Identifying the Common Factor
When factoring polynomials, the first step is often to find and factor out the common factor from each term. In the polynomial \(x^2 y - 16y\), we observe that the term \(y\) appears in both \(x^2 y\) and \(-16y\). This means \(y\) is a common factor.

To factor it out, you divide each term by \(y\) and express the polynomial as \(y(x^2 - 16)\).
  • Make sure every term in the polynomial is considered to identify any common factors.
  • Remember, factoring out the greatest common factor simplifies the polynomial, making further steps easier.
Mastering the skill of finding common factors is key to simplifying expressions in algebra and sets the stage for more complex factorization techniques.
Understanding the Difference of Squares
The expression \(x^2 - 16\) from our polynomial is an example of the difference of squares. The difference of squares formula is crucial when factoring expressions that look like \(a^2 - b^2\).

Here's why:
  • It can be expressed as the product \((a+b)(a-b)\).
  • In our case, \(a^2\) is \(x^2\) and \(b^2\) is \(16\), which translates to \(4^2\).
Hence, \(x^2 - 16\) factors neatly into \((x+4)(x-4)\). This method provides two binomials whose product is the original expression.

Recognizing patterns like the difference of squares is a powerful tool in algebra, often simplifying the factoring process and revealing hidden solutions.
Achieving Complete Factorization
Complete factorization means breaking down a polynomial into simpler polynomials or its factors completely and neatly. After spotting the common factor \(y\) and applying the difference of squares on \(x^2 - 16\), we can express the original polynomial completely factored as \(y(x+4)(x-4)\).

This process involves:
  • Identifying common factors first to simplify the expression.
  • Using formulas like the difference of squares for remaining terms.
  • Ensuring no further factoring is possible, indicating we've reached the simplest form.

This step assures that the polynomial is expressed in a product of polynomials that can't be further broken down. Complete factorization is very helpful in solving equations, simplifying expressions, and is a fundamental skill in advanced algebra topics.

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Most popular questions from this chapter

In \(27-39,\) factor each polynomial completely. $$ 4 x^{3}-10 x^{2}+6 x $$

In \(3-8,\) write each polynomial as the product of its greatest common monomial factor and a polynomial. $$ x^{3} y^{3}-2 x^{3} y^{2}+x^{2} y^{2} $$

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