Question

solve the equation by factoring.

1. $x^2 + 12x + 35 = 0$
2. $2x^2 - 8x + 6 = 0$
3. $3x^2 - 5x = 2$

find the zero(s) of the function.

1. $f(x) = x^2 - 8x$
2. $f(x) = x^2 + 2x - 8$
3. $f(x) = 4x^2 + 28x + 49$

Explanation:

Problem 9: Solve \( x^2 + 12x + 35 = 0 \) by factoring

Step 1: Find two numbers that multiply to 35 and add to 12.

The numbers are 5 and 7, since \( 5 \times 7 = 35 \) and \( 5 + 7 = 12 \).

Step 2: Factor the quadratic.

\( x^2 + 12x + 35 = (x + 5)(x + 7) \)

Step 3: Set each factor equal to zero and solve for \( x \).

\( x + 5 = 0 \) or \( x + 7 = 0 \)
\( x = -5 \) or \( x = -7 \)

Step 1: Factor out the greatest common factor (GCF) of 2.

\( 2(x^2 - 4x + 3) = 0 \)

Step 2: Factor the quadratic inside the parentheses.

Find two numbers that multiply to 3 and add to -4. The numbers are -1 and -3.
\( x^2 - 4x + 3 = (x - 1)(x - 3) \)
So the factored form is \( 2(x - 1)(x - 3) = 0 \)

Step 3: Set each factor equal to zero and solve for \( x \).

\( x - 1 = 0 \) or \( x - 3 = 0 \)
\( x = 1 \) or \( x = 3 \)

Step 1: Rewrite the equation in standard form (\( ax^2 + bx + c = 0 \)).

Subtract 2 from both sides: \( 3x^2 - 5x - 2 = 0 \)

Step 2: Factor the quadratic.

Find two numbers that multiply to \( 3 \times (-2) = -6 \) and add to -5. The numbers are -6 and 1.
Rewrite the middle term: \( 3x^2 - 6x + x - 2 = 0 \)
Group and factor: \( 3x(x - 2) + 1(x - 2) = 0 \)
\( (3x + 1)(x - 2) = 0 \)

Step 3: Set each factor equal to zero and solve for \( x \).

\( 3x + 1 = 0 \) or \( x - 2 = 0 \)
\( x = -\frac{1}{3} \) or \( x = 2 \)

Answer:

\( x = -5 \), \( x = -7 \)

Problem 10: Solve \( 2x^2 - 8x + 6 = 0 \) by factoring