Problem 8 - Elementary and Intermediate Algebra

Question

Fill in the blanks. a. \(36 y^{2}-49 m^{4}=(\quad)^{2}-(\quad)^{2}\) b. \(125 h^{3}-27 k^{6}=(\quad)^{3}-(\quad)^{3}\)

Step-by-Step Solution

Answer

a. \((6y)^2 - (7m^2)^2\); b. \((5h)^3 - (3k^2)^3\).

Step 1: Identify the Expression Type for Part (a)

The expression given is \(36 y^2 - 49 m^4\). We can see that both terms are perfect squares: \(36 y^2 = (6y)^2\) and \(49 m^4 = (7m^2)^2\). Thus, this expression can be expressed as a difference of squares.

Step 2: Express as Difference of Squares for Part (a)

Now that we have identified each term as a perfect square, the expression \(36 y^2 - 49 m^4\) can be rewritten as \((6y)^2 - (7m^2)^2\). Thus \((6y)\) and \((7m^2)\) are the terms that fill the blanks.

Step 3: Identify the Expression Type for Part (b)

The expression given is \(125 h^3 - 27 k^6\). Each term is a perfect cube: \(125 h^3 = (5h)^3\) and \(27 k^6 = (3k^2)^3\). So, this expression can be expressed as a difference of cubes.

Step 4: Express as Difference of Cubes for Part (b)

Both terms are perfect cubes, so the expression \(125 h^3 - 27 k^6\) can be rewritten as \((5h)^3 - (3k^2)^3\). Thus \((5h)\) and \((3k^2)\) are the terms that fill the blanks.

Key Concepts

Difference of Squares

The Difference of Squares is a concept in algebra where an expression takes the form of \[a^2 - b^2\]. This expression can be factored into the product of two binomials: \[(a-b)(a+b)\]. This property is based on the foundational identity that the middle terms cancel out, leaving only the subtraction between the two squares.

A practical example of this would be:

• Take the expression \[36 y^2 - 49 m^4\].
• We identify the first term, \[36 y^2\], as \[(6y)^2\], and the second term, \[49 m^4\], as \[(7m^2)^2\].
• By applying the difference of squares identity, we rewrite it as \[(6y - 7m^2)(6y + 7m^2)\].

This simple yet powerful factorization method is valuable in simplifying expressions and solving equations.

Difference of Cubes

The Difference of Cubes involves expressions of the form \[a^3 - b^3\]. It can be factored using the identity: \[(a-b)(a^2 + ab + b^2)\]. Unlike the difference of squares, the factorization includes three terms in one of the factors.

Here’s how it is applied:

• Consider the expression \[125 h^3 - 27 k^6\].
• This expression can be seen as \[(5h)^3 - (3k^2)^3\].
• Following the difference of cubes rule, it becomes \[(5h - 3k^2)(25h^2 + 15hk^2 + 9k^4)\].

Using this rule helps in simplifying expressions and tackling polynomial equations, providing deeper insights into algebraic structures.

Perfect Squares

A Perfect Square is an expression that results from squaring a binomial. It generally takes the form of \[(a+b)^2 = a^2 + 2ab + b^2\] or \[(a-b)^2 = a^2 - 2ab + b^2\]. Recognizing perfect squares is essential in both expansion and factorization exercises.

Consider the expression \[36 y^2\]. Since \[36 = 6^2\], we identify this as a perfect square \[(6y)^2\]. Seeing perfect squares allows you to simplify even complex expressions or equations, especially when they appear in quadratic equations.

Perfect Cubes

Perfect Cubes refer to expressions derived from raising a number or a term to the power of three. This typically appears as \[a^3\]. A perfect cube is helpful in recognizing patterns in equations, particularly regarding cube roots and factorization.

For example:

• In the expression \[125 h^3\], we have \[(5h)^3\], thus understanding it as a perfect cube.
• Similarly, \[27 k^6\] is \[(3k^2)^3\].

Understanding perfect cubes unlocks the potential for more advanced polynomial manipulation, including solving equations via cube roots.