Question

5.3 - online quiz - solving quadratic equations by fac
started: dec 12 at 6:18pm
quiz instructions
question 3
solve the following equation by factoring.
$6x^2 + 23x + 10 = 0$
$\circ$ $x = 10/3, x = 1/2$
$\circ$ $x = 4, x = 1/3$
$\circ$ $x = -10/3, x = -1/2$
$\circ$ $x = -4, x = -1/3$

Explanation:

Step1: Factor the quadratic

We need to factor \(6x^2 + 23x + 10\). We look for two numbers \(a\) and \(b\) such that \(a\times b = 6\times10 = 60\) and \(a + b = 23\). The numbers are \(20\) and \(3\).
Rewrite the middle term: \(6x^2 + 20x + 3x + 10 = 0\)
Group the terms: \((6x^2 + 20x)+(3x + 10)=0\)
Factor out common factors: \(2x(3x + 10)+1(3x + 10)=0\)
Now, factor out \((3x + 10)\): \((2x + 1)(3x + 10)=0\)

Step2: Solve for \(x\)

Set each factor equal to zero:

• \(2x + 1 = 0\) gives \(2x=-1\) so \(x = -\frac{1}{2}\)
• \(3x + 10 = 0\) gives \(3x=-10\) so \(x = -\frac{10}{3}\)

Answer:

\(x = -\frac{10}{3}, x = -\frac{1}{2}\) (corresponding to the option: \(x = -10/3, x = -1/2\))