/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Factor the polynomial. $$x^{4}... [FREE SOLUTION] | 91Ó°ÊÓ

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Factor the polynomial. $$x^{4}-4 x^{2}$$

Short Answer

Expert verified
The polynomial factors as \(x^2(x-2)(x+2)\).

Step by step solution

01

Identify Common Factors

The polynomial given is \(x^4 - 4x^2\). Start by looking for common factors in each term. Both terms have \(x^2\) as a factor, so factor \(x^2\) out of each term: \(x^4 = x^2 \cdot x^2\) and \(4x^2 = x^2 \cdot 4\). Thus, we can write: \(x^4 - 4x^2 = x^2(x^2 - 4)\).
02

Recognize the Difference of Squares

Notice that the expression inside the parentheses, \(x^2 - 4\), is a difference of squares. Recall that \(a^2 - b^2 = (a-b)(a+b)\). Here, \(x^2 - 4\) can be rewritten as \((x)^2 - (2)^2\).
03

Factor the Difference of Squares

Apply the difference of squares formula to factor \(x^2 - 4\). Replace \((x)^2 - (2)^2\) with \((x-2)(x+2)\). This gives us: \(x^2(x-2)(x+2)\).
04

Final Factored Form

The complete factorization of the original polynomial \(x^4 - 4x^2\) is \(x^2(x-2)(x+2)\).

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Key Concepts

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

Difference of squares
When factoring polynomials, especially quadratic ones, recognizing a difference of squares can be immensely helpful. A difference of squares is a special case where two square numbers are subtracted. The expression takes the form \(a^2 - b^2\). The beauty of this pattern is that it can always be factored into \((a - b)(a + b)\).

Why does this work? Multiplying \((a - b)(a + b)\) results in:
  • \((a-b)(a+b) = a^2 + ab - ab - b^2\)
The \(+ab\) and \(-ab\) terms cancel each other out, leaving us with \(a^2 - b^2\). Recognizing this pattern quickly allows you to convert a polynomial into a product of two simpler binomials.

In the context of our exercise, the expression \(x^2 - 4\) is a classic example. Here, \(x^2\) is a square term, and \(4\) is \((2)^2\), making it a difference of squares. Therefore, it can be factored as \((x-2)(x+2)\). This step is critical for breaking down more complex polynomials into easier parts.
Common factors
Finding common factors is often the first step in simplifying any polynomial. A common factor is a number or expression that divides each term in the polynomial without a remainder. It acts like a "glue" that binds terms together.

In our example, \(x^4 - 4x^2\), the common factor between both terms is \(x^2\). Each term in the expression shares at least two \(x\) factors (i.e., \(x^2\)). Recognizing this, we factor \(x^2\) out of each term:
  • \(x^4 = x^2 \cdot x^2\)
  • \(4x^2 = x^2 \cdot 4\)
By extracting \(x^2\), you rewrite the polynomial: \(x^4 - 4x^2 = x^2(x^2 - 4)\). This simplification sets the stage for further factoring, such as recognizing and applying the difference of squares method. Always start by checking for common factors to reduce complexity early on.
Factoring polynomials
Factoring polynomials is like finding the pieces of a puzzle to express a polynomial as a product of simpler expressions. This skill is foundational in algebra and helps in solving equations, simplifying expressions, and understanding functions.

When approaching a problem like \(x^4 - 4x^2\), begin by examining each term individually. First, check for common factors to simplify the expression. In our example, factoring out \(x^2\) simplifies the polynomial to \(x^2(x^2 - 4)\).

Next, use pattern recognition techniques, such as identifying a difference of squares. This allows you to break down even more complex expressions. The expression \(x^2 - 4\) is a difference of squares and can be factored into \((x-2)(x+2)\).

So, the entire polynomial \(x^4 - 4x^2\) is factored into \(x^2(x-2)(x+2)\). Understanding these steps and recognizing these patterns makes the process of factoring much more manageable, providing clarity and efficiency in solving polynomial equations.

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Most popular questions from this chapter

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