Question

factor: $x^2 - 5x - 14$

○ cannot be factored
○ (x+14)(x-1)
○ (x-14)(x+1)
○ (x-7)(x+2)
○ (x+7)(x-2)

question 6
1 pts
which equation has the following solutions for x?

$x = -7$ or $x = \frac{1}{3}$

○ $3x^2 + 20x - 7 = 0$
○ $2x^2 + 12x - 4 = 0$
○ $x^2 + \frac{1}{3}x - 7 = 0$
○ $x^2 - 7x + \frac{1}{3} = 0$

Explanation:

First Question (Factoring \(x^2 - 5x - 14\))

Step1: Recall factoring form

For a quadratic \(x^2 + bx + c\), we need two numbers that multiply to \(c\) and add to \(b\). Here, \(c=-14\) and \(b = -5\).

Step2: Find two numbers

We need two numbers that multiply to \(-14\) and add to \(-5\). The numbers are \(-7\) and \(2\) (since \(-7\times2=-14\) and \(-7 + 2=-5\)).

Step3: Write the factors

So, \(x^2 - 5x - 14=(x - 7)(x + 2)\).

Step1: Use factored form

If \(x=-7\) or \(x=\frac{1}{3}\), then the factored form is \((x + 7)(3x - 1)=0\) (multiplying the second factor by 3 to eliminate the fraction).

Step2: Expand the factored form

\((x + 7)(3x - 1)=3x^2 - x + 21x - 7=3x^2 + 20x - 7\). So the equation is \(3x^2 + 20x - 7 = 0\).

Answer:

D. \((x - 7)(x + 2)\)

Second Question (Equation with solutions \(x=-7\) or \(x=\frac{1}{3}\))