Question

1. $x^2 - 7x + 6$

factored form:
solution(s):

1. $x^2 - 6x + 9$

factored form:
solution(s):

1. $x^2 - 5x + 6$

factored form:
solution(s):

1. $x^2 - 4x + 4$

factored form:
solution(s):

1. $x^2 - 7x - 8$

factored form:
solution(s):

1. $x^2 + 4x - 32$

factored form:
solution(s):

Explanation:

Problem 9: \( x^2 - 7x + 6 \)

Step 1: Factor the quadratic

We need two numbers that multiply to \( 6 \) and add to \( -7 \). The numbers are \( -1 \) and \( -6 \).
So, \( x^2 - 7x + 6=(x - 1)(x - 6) \)

Step 2: Find the solutions

Set each factor equal to zero:
\( x - 1 = 0 \) gives \( x = 1 \)
\( x - 6 = 0 \) gives \( x = 6 \)

Step 1: Factor the quadratic

This is a perfect square trinomial. \( x^2 - 6x + 9=(x - 3)^2 \) (since \( (a - b)^2=a^2 - 2ab + b^2 \), here \( a = x \), \( b = 3 \), \( 2ab = 6x \))

Step 2: Find the solutions

Set \( (x - 3)^2 = 0 \), then \( x - 3 = 0 \), so \( x = 3 \) (double root)

Step 1: Factor the quadratic

We need two numbers that multiply to \( 6 \) and add to \( -5 \). The numbers are \( -2 \) and \( -3 \).
So, \( x^2 - 5x + 6=(x - 2)(x - 3) \)

Step 2: Find the solutions

Set each factor equal to zero:
\( x - 2 = 0 \) gives \( x = 2 \)
\( x - 3 = 0 \) gives \( x = 3 \)

Answer:

Factored Form: \( (x - 1)(x - 6) \)
Solution(s): \( x = 1, x = 6 \)

Problem 10: \( x^2 - 6x + 9 \)