Question

finding zeros by factoring

1. $y = x^2 + 6x + 9$

$y =$

1. $y = x^2 + x - 20$

$y =$

1. $y = x^2 + 14x + 33$

$y =$

1. $y = x^2 + 8x + 12$
2. $y = x^2 + 5x - 14$
3. $y = x^2 - 9x + 20$
4. $y = x^2 - x - 56$
5. $y = x^2 + 2x - 80$

Explanation:

Let's solve these quadratic equations by factoring to find their zeros. We'll take each equation one by one.

Problem 1: \( y = x^2 + 6x + 9 \)

Step 1: Identify the form

This is a perfect square trinomial since \( 9 = 3^2 \) and \( 6x = 2 \times x \times 3 \).
The formula for a perfect square trinomial is \( a^2 + 2ab + b^2 = (a + b)^2 \). Here, \( a = x \) and \( b = 3 \).

Step 2: Factor the trinomial

\( x^2 + 6x + 9 = (x + 3)^2 \)

Step 3: Find the zeros

Set \( y = 0 \):
\( (x + 3)^2 = 0 \)
Taking the square root of both sides:
\( x + 3 = 0 \)
\( x = -3 \) (double root)

Step 1: Find two numbers

We need two numbers that multiply to \( -20 \) (the constant term) and add up to \( 1 \) (the coefficient of \( x \)).
The numbers are \( 5 \) and \( -4 \) because \( 5 \times (-4) = -20 \) and \( 5 + (-4) = 1 \).

Step 2: Factor the trinomial

\( x^2 + x - 20 = (x + 5)(x - 4) \)

Step 3: Find the zeros

Set \( y = 0 \):
\( (x + 5)(x - 4) = 0 \)
Using the zero - product property:
\( x + 5 = 0 \) or \( x - 4 = 0 \)
\( x = -5 \) or \( x = 4 \)

Step 1: Find two numbers

We need two numbers that multiply to \( 33 \) and add up to \( 14 \).
The numbers are \( 11 \) and \( 3 \) because \( 11\times3 = 33 \) and \( 11 + 3=14 \).

Step 2: Factor the trinomial

\( x^2 + 14x + 33=(x + 11)(x + 3) \)

Step 3: Find the zeros

Set \( y = 0 \):
\( (x + 11)(x + 3)=0 \)
Using the zero - product property:
\( x+11 = 0 \) or \( x + 3=0 \)
\( x=-11 \) or \( x=-3 \)

Answer:

Factored form: \( y = (x + 3)^2 \)
Zeros: \( x = -3 \) (with multiplicity 2)

Problem 2: \( y = x^2 + x - 20 \)