Question

finding zeros by factoring (a>1)

1. $y = 2x^2 + x - 10$
2. $y = 2x^2 + 11x + 5$
3. $y = 3x^2 + x - 4$
4. $y = 5x^2 + 17x + 6$

Explanation:

Problem 1: \( y = 2x^2 + x - 10 \)

Step 1: Factor the quadratic

We need to find two numbers that multiply to \( 2 \times (-10) = -20 \) and add up to \( 1 \). The numbers are \( 5 \) and \( -4 \).
Rewrite the middle term: \( 2x^2 + 5x - 4x - 10 \)
Group the terms: \( (2x^2 + 5x) + (-4x - 10) \)
Factor out the GCF from each group: \( x(2x + 5) - 2(2x + 5) \)
Factor out the common binomial: \( (2x + 5)(x - 2) \)

Step 2: Set each factor to zero

\( 2x + 5 = 0 \) or \( x - 2 = 0 \)
Solve for \( x \):
For \( 2x + 5 = 0 \), \( 2x = -5 \), so \( x = -\frac{5}{2} \)
For \( x - 2 = 0 \), \( x = 2 \)

Step 1: Factor the quadratic

Find two numbers that multiply to \( 3 \times (-4) = -12 \) and add up to \( 1 \). The numbers are \( 4 \) and \( -3 \).
Rewrite the middle term: \( 3x^2 + 4x - 3x - 4 \)
Group the terms: \( (3x^2 + 4x) + (-3x - 4) \)
Factor out the GCF from each group: \( x(3x + 4) - 1(3x + 4) \)
Factor out the common binomial: \( (3x + 4)(x - 1) \)

Step 2: Set each factor to zero

\( 3x + 4 = 0 \) or \( x - 1 = 0 \)
Solve for \( x \):
For \( 3x + 4 = 0 \), \( 3x = -4 \), so \( x = -\frac{4}{3} \)
For \( x - 1 = 0 \), \( x = 1 \)

Step 1: Factor the quadratic

Find two numbers that multiply to \( 2 \times 5 = 10 \) and add up to \( 11 \). The numbers are \( 10 \) and \( 1 \).
Rewrite the middle term: \( 2x^2 + 10x + x + 5 \)
Group the terms: \( (2x^2 + 10x) + (x + 5) \)
Factor out the GCF from each group: \( 2x(x + 5) + 1(x + 5) \)
Factor out the common binomial: \( (2x + 1)(x + 5) \)

Step 2: Set each factor to zero

\( 2x + 1 = 0 \) or \( x + 5 = 0 \)
Solve for \( x \):
For \( 2x + 1 = 0 \), \( 2x = -1 \), so \( x = -\frac{1}{2} \)
For \( x + 5 = 0 \), \( x = -5 \)

Answer:

The zeros are \( x = -\frac{5}{2} \) and \( x = 2 \)

Problem 2: \( y = 3x^2 + x - 4 \)