Question

3-3 day 1 hw: polynomial identities
factor each completely using difference of squares.

1. ( x^2 - 9 )

2)

1. ( 9a^2 - 16 )

4)
factor each completely using difference of cubes.

1. ( m^3 - 125 )

6)

1. ( 27x^3 - 1 )

8
factor each completely using sum of cubes.

1. ( x^3 + 8 )

1

Explanation:

Problem 1: \( x^2 - 9 \)

Step 1: Recall Difference of Squares Formula

The difference of squares formula is \( a^2 - b^2 = (a - b)(a + b) \). For \( x^2 - 9 \), we can rewrite \( 9 \) as \( 3^2 \), so \( a = x \) and \( b = 3 \).

Step 2: Apply the Formula

Substitute \( a = x \) and \( b = 3 \) into the formula: \( x^2 - 3^2 = (x - 3)(x + 3) \).

Step 1: Identify Squares

Rewrite \( 9a^2 \) as \( (3a)^2 \) and \( 16 \) as \( 4^2 \). Now the expression is \( (3a)^2 - 4^2 \), which fits the difference of squares form \( a^2 - b^2 \) with \( a = 3a \) and \( b = 4 \).

Step 2: Apply Difference of Squares

Using \( a^2 - b^2 = (a - b)(a + b) \), we get \( (3a - 4)(3a + 4) \).

Step 1: Recall Difference of Cubes Formula

The difference of cubes formula is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Rewrite \( 125 \) as \( 5^3 \), so \( a = m \) and \( b = 5 \).

Step 2: Apply the Formula

Substitute \( a = m \) and \( b = 5 \) into the formula: \( m^3 - 5^3 = (m - 5)(m^2 + 5m + 25) \).

Answer:

\( (x - 3)(x + 3) \)

Problem 3: \( 9a^2 - 16 \)