Question

factor. then solve for the values of x.
1
$28x^2 - 58x + 24 = 0$
2
$20x^3 + 11x^2 = 3x$

Explanation:

Problem 1: \(28x^2 - 58x + 24 = 0\)

Step 1: Simplify the equation

First, we can divide the entire equation by 2 to simplify it:
\(14x^2 - 29x + 12 = 0\)

Step 2: Factor the quadratic

We need to find two numbers that multiply to \(14\times12 = 168\) and add up to \(-29\). The numbers are \(-8\) and \(-21\).
Rewrite the middle term:
\(14x^2 - 21x - 8x + 12 = 0\)
Factor by grouping:
\(7x(2x - 3) - 4(2x - 3) = 0\)
\((7x - 4)(2x - 3) = 0\)

Step 3: Solve for \(x\)

Set each factor equal to zero:
\(7x - 4 = 0\) or \(2x - 3 = 0\)
For \(7x - 4 = 0\), we get \(x = \frac{4}{7}\)
For \(2x - 3 = 0\), we get \(x = \frac{3}{2}\)

Step 1: Move all terms to one side

Subtract \(3x\) from both sides to get:
\(20x^3 + 11x^2 - 3x = 0\)

Step 2: Factor out the GCF

Factor out \(x\) from each term:
\(x(20x^2 + 11x - 3) = 0\)

Step 3: Factor the quadratic

We need to find two numbers that multiply to \(20\times(-3) = -60\) and add up to \(11\). The numbers are \(15\) and \(-4\).
Rewrite the middle term:
\(20x^2 + 15x - 4x - 3 = 0\)
Factor by grouping:
\(5x(4x + 3) - 1(4x + 3) = 0\)
\((5x - 1)(4x + 3) = 0\)

Step 4: Solve for \(x\)

Set each factor equal to zero:
\(x = 0\) or \(5x - 1 = 0\) or \(4x + 3 = 0\)
For \(5x - 1 = 0\), we get \(x = \frac{1}{5}\)
For \(4x + 3 = 0\), we get \(x = -\frac{3}{4}\)

Answer:

\(x = \frac{4}{7}\) or \(x = \frac{3}{2}\)

Problem 2: \(20x^3 + 11x^2 = 3x\)