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Chapter 5: Problem 89

Factor completely. $$ (y-1)^{4}-(y-1)^{2} $$

Short Answer

Expert verified
(y-1)^{2} (y-2) y

Step by step solution

01

Identify Common Factor

Notice that both terms \( (y-1)^4 \) and \( (y-1)^2 \) contain the factor \( (y-1)^2 \). Factor out \( (y-1)^2 \):\[ (y-1)^{4} - (y-1)^{2} = (y-1)^2 \big( (y-1)^2 - 1 \big) \]
02

Factor Difference of Squares

Recognize that \( (y-1)^2 - 1 \) is a difference of squares. It can be factored as:\[ (y-1)^2 - 1 = ((y-1) - 1)((y-1) + 1) \]Substitute into the equation:\[ (y-1)^2 \big( (y-1) - 1 \big) \big( (y-1) + 1 \big) \]
03

Simplify the Factors

Simplify the expressions inside the parentheses:\[ (y-1) - 1 = y - 2 \]\[ (y-1) + 1 = y \]Thus, the equation is:\[ (y-1)^{2} (y-2)(y) \]

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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 special pattern in algebra that makes factoring easier. It follows the formula: \[ a^2 - b^2 = (a - b)(a + b) \]This tells us that if you have two squared terms separated by a minus sign, they can be factored into the product of the sum and difference of the terms inside the squares. For example, in our exercise, after factoring out the common factor, we had\[ (y-1)^2 - 1 \]Notice that it fits the difference of squares pattern where \[ a = (y-1) \] and \[ b = 1 \].Applying the difference of squares formula, it became\[ ((y-1) - 1)((y-1) + 1) \]This simplified to\[ (y-2)(y) \].Knowing this pattern can save a lot of time in factoring polynomials.
Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial into simpler products called factors. These factors multiply together to give the original polynomial. For example, consider the polynomial\[ (y-1)^4 - (y-1)^2 \]in our exercise. We broke it down into simpler factors:\[ (y-1)^2 ((y-1) - 1)((y-1) + 1) \]This method often reveals simpler expressions and is essential for solving polynomial equations. By rewriting polynomials in factored form, you can more easily set each factor to zero and solve for the variable. Factorization helps uncover solutions that might be hidden in a more complex original polynomial.
Common Factors
When factoring polynomials, the first step is often to identify any common factors among the terms. A common factor is a term that appears in each part of the polynomial. In our problem,\[ (y-1)^4 - (y-1)^2 \]we saw that \[ (y-1)^2 \] is a common factor in both terms.We factored it out to simplify the expression:\[ (y-1)^2 ((y-1)^2 - 1) \]Factoring out common factors can simplify the polynomial and make other factoring techniques, like the difference of squares, more apparent. Always begin by checking for and factoring out any common factors. This can make the remaining problem much more manageable.

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