MHB Algebraic Expressions Simplified

[loraboiago]

How does 3x or (x-5) equal 0 in the statement 3x(x-5)=0? I don't understand the logic behind it. Thank you!

 

[MarkFL]

If you have the statement:

$$a\cdot b=0$$ where $$a\ne b$$

Then the only way it can be true is if either $$a=0$$ or $$b=0$$. This is called the zero-factor property.

 

[loraboiago]

If you have the statement:

$$a\cdot b=0$$ where $$a\ne b$$

Then the only way it can be true is if either $$a=0$$ or $$b=0$$. This is called the zero-factor property.

Thank you Mark for the quick and helpful response. The answer to this question went on to explain "3x(x-5)=0 provides an equation in which at least one of the expressions 3x or (x-5) is equal to 0. That translates into two possible values for x: 0 and 5."

I understand how one can equal 0 (thanks to you!), but how do I calculate the other possible value as being 5?

 

[MarkFL]

Thank you Mark for the quick and helpful response. The answer to this question went on to explain "3x(x-5)=0 provides an equation in which at least one of the expressions 3x or (x-5) is equal to 0. That translates into two possible values for x: 0 and 5."

I understand how one can equal 0 (thanks to you!), but how do I calculate the other possible value as being 5?

I would look at it as 3 factors being equal to zero:

$$3\cdot x\cdot(x-5)=0$$

Now, set all factors involving $x$ equal to zero, and then solve for $x$ in each equation:

$$x=0$$

$$x-5=0$$

The solutions to these equations will give you the solutions to the original equation.

 

[loraboiago]

I would look at it as 3 factors being equal to zero:

$$3\cdot x\cdot(x-5)=0$$

Now, set all factors involving $x$ equal to zero, and then solve for $x$ in each equation:

$$x=0$$

$$x-5=0$$

The solutions to these equations will give you the solutions to the original equation.

Ah got it! You are awesome. Thank you :)