Question

How do you solve $\left( 2x+8 \right)\left( 3x-6 \right)=0$ using any method?

Answer 1

Hint: In the left hand side of the given equation, we can see that we have a quadratic polynomial into factored form, and expressed as a multiplication of two linear factors, which are $\left( 2x+8 \right)$ and $\left( 3x-6 \right)$. In the first factor $\left( 2x+8 \right)$ we can see that $2$ is common to both the terms and so we can take it out. Also, in the second factor $\left( 3x-6 \right)$ we can see that $3$ is common to both the terms and we can take it out. So we will get a multiplication of two simplified linear factors on the left hand side of the equation. Using the zero product rule we can equate each of the factors to zero, which will give us the two solutions of the given equation.

Complete step-by-step answer:
The equation to be solved is given in the question as
$\left( 2x+8 \right)\left( 3x-6 \right)=0$
We simplify the above equation by taking out $2$ from the first factor $\left( 2x+8 \right)$ to get
$\Rightarrow 2\left( x+4 \right)\left( 3x-6 \right)=0$
Similarly we take out $3$ from the second factor $\left( 3x-6 \right)$ to get
\[\begin{align}
  & \Rightarrow 2\left( x+4 \right)3\left( x-2 \right)=0 \\
 & \Rightarrow 6\left( x+4 \right)\left( x-2 \right)=0 \\
\end{align}\]
Dividing both sides of the above equation by $6$ we get
$\Rightarrow \left( x+4 \right)\left( x-2 \right)=0$
Now, we know that if the product of two terms is equal to zero, then at least one of the two must be equal to zero. Since both of the terms $\left( x+4 \right)$ and \[\left( x-2 \right)\] are variables, so both of them can be zero. So we equate both of these to zero one by one to get the equations
\[\Rightarrow \left( x-2 \right)=0\]
\[\Rightarrow \left( x+4 \right)=0\]
On solving the above equations, we get the solutions as $x=2,-4$.
Hence, the solutions of the given equation $\left( 2x+8 \right)\left( 3x-6 \right)=0$ are $x=2,x=-4$.

Note: We can also solve the given equation by expanding the left hand side by multiplying the two linear factors to obtain a quadratic equation in the standard form. Then, using the quadratic formula or the middle split technique, we can obtain the final solutions, which will be the same as in the above solution.