Problem 138
Question
Factor completely. $$5 y^{5}-5 y^{4}-20 y^{3}+20 y^{2}$$
Step-by-Step Solution
Verified Answer
The completely factored form of the given polynomial \(5y^{5} - 5y^{4} - 20y^{3} + 20y^{2}\) is \(5y^{2}(y + 2)(y - 2)(y - 1)\)
1Step 1: Look for Common Factors
First, we examine the given polynomial, \(5y^{5} - 5y^{4} - 20y^{3} + 20y^{2}\). We can recognize that each individual term in the expression has a common factor of \(5y^{2}\).
2Step 2: Factor Out the Common Factors
We then factor this common factor out of each term, which leaves: \(5y^{2}(y^{3} - y^{2} - 4y + 4)\).
3Step 3: Factor the Remaining Polynomial
Next, we continue factoring the remaining polynomial \(y^3 - y^2 - 4y + 4\). This can be factored by grouping, a common method for factorising four term polynomials. By grouping the terms and factoring, the polynomial becomes: \((y^2 - 4)(y - 1)\).
4Step 4: Completing the Factoring
Finally, \(y^2 - 4\) is a difference of squares and can be factored into \((y + 2)(y - 2)\). Thus, the given polynomial can be completely factored as: \(5y^{2}(y + 2)(y - 2)(y - 1)\).
Key Concepts
Common FactorDifference of SquaresFactor by Grouping
Common Factor
When tackling polynomials, one of the first and most effective strategies is to identify the **common factor**. This involves finding a number or variable that is present in each term of the polynomial.
Here’s how you can do it:
In our example, the polynomial is given by \(5y^{5} - 5y^{4} - 20y^{3} + 20y^{2}\). In all these terms, both the number 5 and the expression \(y^2\) appear. By factoring out \(5y^2\), we simplify our expression to \(5y^2(y^3 - y^2 - 4y + 4)\). This simplification makes the next steps of factoring much easier. By identifying the common factor at the start, you set a strong foundation for solving any polynomial problem.
Here’s how you can do it:
- Examine each term: Look at the numerical coefficients and variables in every term.
- Identify the greatest factor: Determine the largest factor that is common to all the terms.
- Factor it out: Once you identify the common factor, factor it out of the polynomial, simplifying each term.
In our example, the polynomial is given by \(5y^{5} - 5y^{4} - 20y^{3} + 20y^{2}\). In all these terms, both the number 5 and the expression \(y^2\) appear. By factoring out \(5y^2\), we simplify our expression to \(5y^2(y^3 - y^2 - 4y + 4)\). This simplification makes the next steps of factoring much easier. By identifying the common factor at the start, you set a strong foundation for solving any polynomial problem.
Difference of Squares
A significant method in polynomial factoring is recognizing a **difference of squares**. This occurs when you have an expression of the form \(a^2 - b^2\), which can be factored into \((a + b)(a - b)\).
The beauty of the difference of squares is its simplicity and pattern:
In our factoring problem, after applying the common factor method, we arrive at \(y^2 - 4\). This expression can be seen as \(y^2 - 2^2\), fitting perfectly into the difference of squares pattern.
Thus, \(y^2 - 4\) can be factored into \((y + 2)(y - 2)\), which is crucial for further simplification of the polynomial. Recognizing this pattern helps in efficiently breaking down complex polynomials into simpler factors.
The beauty of the difference of squares is its simplicity and pattern:
- It always follows the structure \(a^2 - b^2\).
- The resulting factors are conjugates, \((a + b)\) and \((a - b)\).
In our factoring problem, after applying the common factor method, we arrive at \(y^2 - 4\). This expression can be seen as \(y^2 - 2^2\), fitting perfectly into the difference of squares pattern.
Thus, \(y^2 - 4\) can be factored into \((y + 2)(y - 2)\), which is crucial for further simplification of the polynomial. Recognizing this pattern helps in efficiently breaking down complex polynomials into simpler factors.
Factor by Grouping
**Factor by grouping** is a useful technique when dealing with polynomials that consist of four or more terms. This method involves rearranging and grouping terms to facilitate factoring.
Here's a step-by-step approach:
In our example, after extracting the common factor, we work with \(y^3 - y^2 - 4y + 4\). We group it as \((y^3 - y^2)\) and \((-4y + 4)\). By factoring each group, we extract \(y^2(y - 1)\) and \(-4(y - 1)\), revealing a common binomial \((y - 1)\).
The expression then becomes \((y - 1)(y^2 - 4)\), paving an easier path to final simplification. Using the factor by grouping method helps in dealing with polynomials having more than three terms by focusing on smaller, factorable sections.
Here's a step-by-step approach:
- Group the terms: Pair the polynomial terms in a way that simplifies the expression. Ideally, each group should have a common factor.
- Factor each group: Take out common factors from each pair.
- Combine and simplify: If done correctly, a common binomial factor should emerge, which can then be factored out.
In our example, after extracting the common factor, we work with \(y^3 - y^2 - 4y + 4\). We group it as \((y^3 - y^2)\) and \((-4y + 4)\). By factoring each group, we extract \(y^2(y - 1)\) and \(-4(y - 1)\), revealing a common binomial \((y - 1)\).
The expression then becomes \((y - 1)(y^2 - 4)\), paving an easier path to final simplification. Using the factor by grouping method helps in dealing with polynomials having more than three terms by focusing on smaller, factorable sections.