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Chapter 5: Problem 98

Factor each expression. $$ 1-y^{6} $$

Short Answer

Expert verified
The expression factors to \((1-y)(1+y+y^2)(1+y)(1-y+y^2)\).

Step by step solution

01

Recognize the pattern

The expression is of the form 1 minus a perfect square, which is a difference of squares. It can be expressed as \( (1)^2 - (y^3)^2 \).
02

Apply difference of squares formula

The difference of squares formula is \( a^2 - b^2 = (a-b)(a+b) \). For this expression, \( a = 1 \) and \( b = y^3 \).
03

Factor using the formula

Substitute \( a \) and \( b \) into the formula: \[ 1-y^6 = (1-y^3)(1+y^3) \].
04

Factor further using difference of cubes (optional)

The expression \( 1-y^3 \) can be further factored using the difference of cubes formula: \( (1-y^3) = (1-y)(1+y+y^2) \). The expression \( 1+y^3 \) can also be factored using the sum of cubes: \( (1+y^3) = (1+y)(1-y+y^2) \).
05

Write final expression

Combining the factors, the fully factored expression is: \[ (1-y)(1+y+y^2)(1+y)(1-y+y^2) \].

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Key Concepts

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

Difference of Squares
The difference of squares is a common technique used in algebra to factor expressions. It is based on the identity \( a^2 - b^2 = (a-b)(a+b) \). In simple terms, whenever you have two terms each raised to a power of two, separated by a minus sign, you can factor them using this formula.For instance, consider the expression \( 1 - y^6 \). We can view this as a perfect square problem because we recognize it as \( (1)^2 - (y^3)^2 \). Here:
  • \( a = 1 \)
  • \( b = y^3 \)
Using the difference of squares formula, it can be factored into \( (1-y^3)(1+y^3) \). The simplicity of this method makes it a favorite when dealing with polynomial expressions in algebra.
Difference of Cubes
The difference of cubes formula allows us to factor expressions of the form \( a^3 - b^3 \). The formula used is \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Understanding this pattern aids in breaking down more complex cubic expressions into simpler, manageable parts.Digging into our specific example, when factoring \( 1 - y^6 \), we initially get \( 1-y^3 \) from the difference of squares. Applying the difference of cubes formula, with:
  • \( a = 1 \)
  • \( b = y \)
We can factor it as \( (1-y)(1+y+y^2) \). This breakdown helps further simplify your polynomial, making it easier to understand and use in broader mathematical problems.
Sum of Cubes
The sum of cubes is closely related to the difference of cubes. The formula to factor these types of expressions is \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \). Like its counterpart, the sum of cubes helps turn complex cubic expressions into a product of simpler binomials and trinomials.Returning to our example, after applying the difference of squares, we address \( 1 + y^3 \). Using the sum of cubes formula where:
  • \( a = 1 \)
  • \( b = y \)
We can factor it to become \( (1+y)(1-y+y^2) \). This effective technique breaks down the polynomial into a user-friendly format, facilitating better understanding and manipulation in various algebraic applications.

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Most popular questions from this chapter

The area of a square checkerboard is represented by the polynomial \(25 x^{2}-40 x+16 .\) Use factoring to find an expression that represents the length of a side. (IMAGE CANT COPY)

Factor the expression in part a. Then use your answer from part a to give the factorization of the expression in part b. (No new work is necessary!) a. \(c^{3}-\frac{1}{8}\) b. \(c^{3}+\frac{1}{8}\)

Look Alikes . . . a. \(a^{2}-b^{2}\) b. \(a^{3}-b^{3}\)

Factor each expression. $$ x^{2}+20 x+100-9 z^{2} $$

How do you know when a polynomial has been factored completely?
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