Question

what are the solutions to the equation ( 5(x + 12)(x - 7)=0 )?

Explanation:

Step1: Expand the left - hand side

We are given the equation \(7(x + 12)(x-7)=0\). First, we can use the zero - product property. The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). In our case, we can first consider the factors. But we can also expand the left - hand side.
First, expand \((x + 12)(x - 7)\) using the FOIL method: \((x+12)(x - 7)=x\times x+x\times(-7)+12\times x + 12\times(-7)=x^{2}-7x + 12x-84=x^{2}+5x - 84\)
Then the original equation becomes \(7(x^{2}+5x - 84)=0\). Divide both sides of the equation by 7: \(x^{2}+5x - 84=0\)

Step2: Solve the quadratic equation

We can solve the quadratic equation \(x^{2}+5x - 84 = 0\) by factoring. We need to find two numbers that multiply to \(- 84\) and add up to \(5\).
Let's find the factors of \(-84\):
The factors of \(-84\) are: \(-7\) and \(12\) since \(-7\times12=-84\) and \(-7 + 12 = 5\)
So we can factor the quadratic as \((x - 7)(x+12)=0\)
Using the zero - product property:
If \(x - 7=0\), then \(x = 7\)
If \(x + 12=0\), then \(x=-12\)

Wait, but let's check the original equation again. The original equation is \(7(x + 12)(x - 7)=0\). By the zero - product property, we can set each factor equal to zero.
Set \(x + 12=0\), then \(x=-12\)
Set \(x - 7=0\), then \(x = 7\)

But looking at the options, if we made a mistake in expansion, let's go back. Wait, maybe the original equation is \(7(x + 12)(x - 7)=0\), so either \(x+12 = 0\) or \(x - 7=0\) or \(7 = 0\) (but \(7
eq0\)). So \(x=-12\) or \(x = 7\). But looking at the options, if we consider the equation \(7(x + 12)(x - 7)=0\), dividing both sides by \(7\) gives \((x + 12)(x - 7)=0\), so \(x=-12\) or \(x = 7\). But if we made a mistake in the problem statement (maybe a typo), and the equation is \(7(x + 2)(x - 7)=0\) (just a guess for the sake of matching the options), but based on the given equation \(7(x + 12)(x - 7)=0\), the solutions are \(x = 7\) and \(x=-12\).

Wait, maybe the original equation was \(7(x + 2)(x - 7)=0\). Let's check:
If \(7(x + 2)(x - 7)=0\), divide by \(7\): \((x + 2)(x - 7)=0\)
Then \(x+2=0\) or \(x - 7=0\), so \(x=-2\) or \(x = 7\). But this is just a guess.

Wait, let's re - examine the original problem. The user's image shows the equation \(7(x + 12)(x - 7)=0\). So the solutions are found by setting each factor equal to zero:

1. \(x+12 = 0\Rightarrow x=-12\)
2. \(x - 7=0\Rightarrow x = 7\)

But looking at the options, if we assume that there was a typo and the equation is \(7(x + 2)(x - 7)=0\), then:
Divide by \(7\): \((x + 2)(x - 7)=0\)
\(x+2=0\Rightarrow x=-2\); \(x - 7=0\Rightarrow x = 7\)

Alternatively, if the equation is \(7(x + 12)(x+7)=0\), then:
\(x+12 = 0\Rightarrow x=-12\); \(x + 7=0\Rightarrow x=-7\)

But based on the given equation \(7(x + 12)(x - 7)=0\), the solutions are \(x = 7\) and \(x=-12\)

Answer:

The solutions of the equation \(7(x + 12)(x - 7)=0\) are \(x = 7\) and \(x=-12\)