Question

1. ( x^2 + 11x + 30 )
2. ( 6x^2 - 5x + 1 )

(2x - 1)(3x - 1

1. ( 3x^2 + 3x - 6 )

Explanation:

Problem 7: \( x^2 + 11x + 30 \)

Step1: Find two numbers

Find two numbers that multiply to \( 30 \) (the constant term) and add up to \( 11 \) (the coefficient of \( x \)). The numbers are \( 5 \) and \( 6 \) since \( 5 \times 6 = 30 \) and \( 5 + 6 = 11 \).

Step2: Factor the quadratic

Rewrite the middle term using these two numbers: \( x^2 + 5x + 6x + 30 \). Then factor by grouping: \( x(x + 5) + 6(x + 5) = (x + 5)(x + 6) \).

Step1: Find \( ac \) and numbers

For \( ax^2 + bx + c \), \( a = 6 \), \( b = -5 \), \( c = 1 \). \( ac = 6 \times 1 = 6 \). Find two numbers that multiply to \( 6 \) and add to \( -5 \). The numbers are \( -2 \) and \( -3 \) (since \( -2 \times -3 = 6 \) and \( -2 + (-3) = -5 \)).

Step2: Rewrite and factor

Rewrite the middle term: \( 6x^2 - 2x - 3x + 1 \). Factor by grouping: \( 2x(3x - 1) - 1(3x - 1) = (2x - 1)(3x - 1) \).

Step1: Factor out GCF

First, factor out the greatest common factor (GCF) of \( 3 \) from all terms: \( 3(x^2 + x - 2) \).

Step2: Factor the quadratic

For \( x^2 + x - 2 \), find two numbers that multiply to \( -2 \) and add to \( 1 \). The numbers are \( 2 \) and \( -1 \) (since \( 2 \times -1 = -2 \) and \( 2 + (-1) = 1 \)). Rewrite and factor: \( x^2 + 2x - x - 2 = x(x + 2) - 1(x + 2) = (x + 2)(x - 1) \). Then multiply back the GCF: \( 3(x + 2)(x - 1) \).

Answer:

\( (x + 5)(x + 6) \)

Problem 10: \( 6x^2 - 5x + 1 \)