Problem 1
Question
Match each polynomial in Column I with its factored form in Column II. I (a) \(x^{2}+10 x y+25 y^{2}\) (b) \(x^{2}-10 x y+25 y^{2}\) (c) \(x^{2}-25 y^{2}\) (d) \(25 y^{2}-x^{2}\) II A. \((x+5 y)(x-5 y)\) B. \((x+5 y)^{2}\) C. \((x-5 y)^{2}\) D. \((5 y+x)(5 y-x)\)
Step-by-Step Solution
Verified Answer
(a) - B, (b) - C, (c) - A, (d) - D.
1Step 1: Identify Factoring Patterns
Polynomials can often be factored using special factoring formulas such as perfect square trinomials and difference of squares. Perfect square trinomials follow the pattern \( a^2 \pm 2ab + b^2 = (a \pm b)^2 \), while the difference of squares uses \( a^2 - b^2 = (a + b)(a - b) \).
2Step 2: Factor Polynomial (a)
The polynomial \( x^{2}+10xy+25y^{2} \) matches the pattern of a perfect square trinomial, specifically \( (a + b)^2 \). Here, \( a = x \) and \( b = 5y \), as \( 2ab = 10xy \). Therefore, it factors into \((x+5y)^2\).
3Step 3: Factor Polynomial (b)
The polynomial \( x^{2}-10xy+25y^{2} \) is another perfect square trinomial, matching the pattern \( (a - b)^2 \). In this case, \( a = x \) and \( b = 5y \), as \( -2ab = -10xy \). It factors into \((x-5y)^2\).
4Step 4: Factor Polynomial (c)
The polynomial \( x^{2}-25y^{2} \) fits the difference of squares pattern \( a^2 - b^2 \), with \( a = x \) and \( b = 5y \). This factors into \((x+5y)(x-5y)\).
5Step 5: Factor Polynomial (d)
The polynomial \( 25y^{2}-x^{2} \) is a difference of squares as well, written in reverse order. It can be expressed as \((5y)^2-x^2\), which factors into \((5y+x)(5y-x)\).
6Step 6: Match Columns
Using the factored forms from previous steps:- (a) matches with II: B, because (a) factors into \((x+5y)^2\).- (b) matches with II: C, because (b) factors into \((x-5y)^2\).- (c) matches with II: A, because (c) factors into \((x+5y)(x-5y)\).- (d) matches with II: D, because (d) factors into \((5y+x)(5y-x)\).
Key Concepts
Perfect Square TrinomialsDifference of SquaresFactoring Techniques
Perfect Square Trinomials
Perfect square trinomials are a special type of polynomial that follows a specific pattern, making them straightforward to factor. They are usually of the form \( a^2 \pm 2ab + b^2 \). Here's what this means:
In our exercise, polynomials like \( x^2 + 10xy + 25y^2 \) become easily factored because it matches \( (a+b)^2 \), with \( a = x \) and \( b = 5y \), leading to \((x+5y)^2\).
- The expression \( a^2 \) represents the square of term \( a \).
- \( 2ab \) is twice the product of two terms, \( a \) and \( b \).
- \( b^2 \) is the square of the term \( b \).
- If the middle term is positive, as in \( a^2 + 2ab + b^2 \), it can be factored as \((a+b)^2\).
- If the middle term is negative, such as \( a^2 - 2ab + b^2 \), it factors into \((a-b)^2\).
In our exercise, polynomials like \( x^2 + 10xy + 25y^2 \) become easily factored because it matches \( (a+b)^2 \), with \( a = x \) and \( b = 5y \), leading to \((x+5y)^2\).
Difference of Squares
The difference of squares is another common pattern in polynomials, recognizable by its simple structure \( a^2 - b^2 \). This pattern is defined by two essential characteristics:
For instance, with \( x^2 - 25y^2 \), notice that \( x^2 \) is a square, and \( 25y^2 \) is also a square \((5y)^2\). Hence, using the difference of squares, it simplifies to \((x+5y)(x-5y)\).
This method simplifies solving polynomial equations significantly, as it breaks them into manageable parts.
- Two terms that are squares.
- A subtraction operation between these two squared terms.
- \( a^2 - b^2 = (a + b)(a - b) \)
For instance, with \( x^2 - 25y^2 \), notice that \( x^2 \) is a square, and \( 25y^2 \) is also a square \((5y)^2\). Hence, using the difference of squares, it simplifies to \((x+5y)(x-5y)\).
This method simplifies solving polynomial equations significantly, as it breaks them into manageable parts.
Factoring Techniques
Factoring techniques are crucial in breaking down polynomials into simpler components. They are methods used to express a polynomial as a product of its factors. Sometimes these techniques involve special formulas such as perfect square trinomials and differences of squares. Here's why these methods are important:
- They simplify complex polynomial expressions.
- Help in solving polynomial equations by making it easier to find zeroes or roots.
- Assist in graphical representation by breaking polynomials into scalable parts.
- Identity special forms like perfect squares and difference of squares immediately.
- Look for common factors in each term.
- Write polynomials in descending power order to ease factorization.
Other exercises in this chapter
Problem 1
Match the rational exponent expression in Exercises \(1-8\) with the equivalent radical expression in \(A-H\). Assume that \(x \neq 0\). A. \(\frac{3}{\sqrt[3]{
View solution Problem 1
Find the domain of each rational expression. $$\frac{x-2}{x+6}$$
View solution Problem 1
Simplify each expression. Leave answers with exponents. $$(-4)^{3} \cdot(-4)^{2}$$
View solution Problem 2
Concept Check Match each expression from Group I with the correct choice from Group II. Choices may be used once, more than once, or not at all. (II) \(\begin{a
View solution