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exit ticket factor each quadratic expression. 1) $x^2 - x - 42$ 2) $4x^…

Question

exit ticket
factor each quadratic expression.

  1. $x^2 - x - 42$
  2. $4x^2 - 81$

Explanation:

Response
Problem 1: Factor \( x^2 - x - 42 \)

Step 1: Find two numbers

We need two numbers that multiply to \( -42 \) (the constant term) and add up to \( -1 \) (the coefficient of \( x \)). Let's list the factor pairs of \( 42 \): \( 1 \) and \( 42 \), \( 2 \) and \( 21 \), \( 3 \) and \( 14 \), \( 6 \) and \( 7 \). We need one positive and one negative number. Let's check: \( -7 \) and \( 6 \) multiply to \( -42 \) (\( -7 \times 6 = -42 \)) and add up to \( -1 \) (\( -7 + 6 = -1 \)).

Step 2: Factor the quadratic

Using these two numbers, we can factor the quadratic as \( (x - 7)(x + 6) \). Let's verify: \( (x - 7)(x + 6) = x^2 + 6x - 7x - 42 = x^2 - x - 42 \), which matches the original expression.

Problem 2: Factor \( 4x^2 - 81 \)

Step 1: Identify the form

Notice that \( 4x^2 - 81 \) is a difference of squares. The difference of squares formula is \( a^2 - b^2 = (a - b)(a + b) \). Here, \( a = 2x \) (since \( (2x)^2 = 4x^2 \)) and \( b = 9 \) (since \( 9^2 = 81 \)).

Step 2: Apply the formula

Using the difference of squares formula, we factor \( 4x^2 - 81 \) as \( (2x - 9)(2x + 9) \). Let's verify: \( (2x - 9)(2x + 9) = (2x)^2 - 9^2 = 4x^2 - 81 \), which matches the original expression.

Answer:

s:

  1. \( \boldsymbol{(x - 7)(x + 6)} \)
  2. \( \boldsymbol{(2x - 9)(2x + 9)} \)