Question

1. solve for the values of x by factoring.

$-3x^2 + 4x = -15$
$\bigcirc$ $x = -3; x = -\frac{3}{5}$
$\bigcirc$ $x = 4; x = -\frac{4}{3}$
$\bigcirc$ $x = -3; x = \frac{5}{3}$
$\bigcirc$ $x = 3; x = -\frac{5}{3}$

Explanation:

Step 1: Rewrite the equation in standard form

Start with the given equation: \( -3x^2 + 4x = -15 \). Add \( 15 \) to both sides to get \( -3x^2 + 4x + 15 = 0 \). Multiply both sides by \( -1 \) to make the coefficient of \( x^2 \) positive: \( 3x^2 - 4x - 15 = 0 \).

Step 2: Factor the quadratic equation

We need to factor \( 3x^2 - 4x - 15 \). We look for two numbers that multiply to \( 3\times(-15) = -45 \) and add up to \( -4 \). The numbers are \( -9 \) and \( 5 \) because \( -9\times5 = -45 \) and \( -9 + 5 = -4 \).
Rewrite the middle term using these numbers: \( 3x^2 - 9x + 5x - 15 = 0 \).
Group the terms: \( (3x^2 - 9x) + (5x - 15) = 0 \).
Factor out the common factors from each group: \( 3x(x - 3) + 5(x - 3) = 0 \).
Now, factor out \( (x - 3) \): \( (3x + 5)(x - 3) = 0 \).

Step 3: Solve for \( x \)

Set each factor equal to zero:

• \( 3x + 5 = 0 \) gives \( 3x = -5 \), so \( x = -\frac{5}{3} \).
• \( x - 3 = 0 \) gives \( x = 3 \).

Answer:

\( x = 3; x = -\frac{5}{3} \)