Question

How do you solve $\left( 3n-2 \right)\left( 4n+1 \right)=0$ by factoring?

Answer 1

Hint: Problems of this type can be solved by applying the property of numbers. If multiplication of two factors is equal to $0$ , then any one of the factors must be $0$ . Equating the factors to $0$ individually and further simplifying we will get the solution of the given equation.

Complete step-by-step solution:
The given equation is
$\left( 3n-2 \right)\left( 4n+1 \right)=0$
As, the quadratic equation is already written in factorized form we can find the solution of the equation by applying property of numbers. We know that if the multiplication of two numbers is equal to $0$ then anyone number among the two has to be $0$ .
So, in our equation either the factor $\left( 3n-2 \right)$ or $\left( 4n+1 \right)$ must be zero. To find the solution we will consider both the cases and equate each factor individually to $0$.
Considering only the first factor to be $0$
$\Rightarrow 3n-2=0$
Adding $2$ to both the sides
$\Rightarrow 3n=2$
Dividing both sides of the equation by $3$ , we get
$\Rightarrow n=\dfrac{2}{3}$
Now, considering only the second term to be $0$
$\Rightarrow 4n+1=0$
Subtracting $1$ from both the sides
$\Rightarrow 4n=-1$
Dividing both sides of the equation by $4$ , we get
 $\Rightarrow n=-\dfrac{1}{4}$
The final solution is all the values that make $\left( 3n-2 \right)\left( 4n+1 \right)=0$ true, which are $n=\dfrac{2}{3},-\dfrac{1}{4}$.

Note: While equating terms to zero we must keep in mind that any of the factors will be equal to $0$ and the other one must not be zero. As in this case, while considering only $\left( 3n-2 \right)$ equal to $0$ , $\left( 4n+1 \right)$ is not equal to zero and vice versa. Also, we must be careful while simplifying the terms to get an appropriate solution to the equation.