Math FoundationsEquations

Zero Product Property

A simple and very effective way to solve equations made of factors (brackets) whose product is zero. You need this method, because a huge number of different equations get reduced to exactly this form.
Key elements:
Connections:
If you can’t solve elementary equations, you definitely won’t solve these either.
Same Action Rule
Solving an Equation
Statistics:
Statement1
Problem1

Brackets and zero

It may seem like once you know how to solve elementary equations, every door is open and now you can handle anything. Not quite. Using the same action rule, we really can do a lot, but there are plenty of equations where you need a trick, not some canned algorithm. Here are a couple of examples:
(x−1)(x−2)=0(x-1)(x-2) = 0
2x(10+x)(2x−1)=0 2x(10+x)(2x-1) = 0
Just expand the brackets and isolate x on one side, what could be easier?! All right, let’s try it:
x2−2x−x+2=0x2−3x+2=0???x^2 - 2x - x + 2 = 0 \\ x^2 - 3x + 2 = 0 \\ \text{???}
2x(20x−10+2x2−x)=02x(2x2+19x−10)=04x3+38x2−20x=0??? 2x(20x - 10 + 2x^2 - x) = 0 \\ 2x(2x^2 + 19x - 10) = 0 \\ 4x^3 + 38x^2 - 20x = 0 \\ \text{???}
Well, shit. That isn’t working. We get too many variables, and on top of that they come in different powers, so combining like terms becomes impossible. We just ran into a whole class of equations that the elementary method can’t handle. So how do we solve them?

How the solution works

These equations are solved very simply! First, look at what they actually look like. They are all just a bunch of factors whose product is guaranteed to be zero:
(x−1)(x−2)=0(x−1)⏟Factor 1⋅(x−2)⏟Factor 2=0(x-1)(x-2) = 0 \\ \underbrace{(x-1)}_{\text{Factor 1}} \cdot \underbrace{(x-2)}_{\text{Factor 2}} = 0
2x(10+x)(2x−1)=0(2x)⏟Factor 1⋅(10+x)⏟Factor 2⋅(2x−1)⏟Factor 3=0 2x(10+x)(2x-1) = 0 \\ \underbrace{(2x)}_{\text{Factor 1}} \cdot \underbrace{(10+x)}_{\text{Factor 2}} \cdot \underbrace{(2x-1)}_{\text{Factor 3}} = 0
We know that anything multiplied by zero gives zero, and it doesn’t matter at all whether that thing is a number, a bracket, or some expression with variables. Zero wipes all of it out:
5⋅0=05 \cdot \red{0} = 0
22⋅0=0 \frac{\sqrt{2}}{2} \cdot \red{0} = 0
0⋅(x−2)=0 \red{0} \cdot (x-2) = 0
We can use this nice fact. We do not have to solve the whole equation at once. We know the result has to be zero, and the left-hand side is nothing but multiplication. It is enough to make at least one factor equal to zero, and it will instantly zero out every other factor, and who gives a damn what numbers they become.
By this logic, the equation (x – 1)(x – 2) = 0 splits into two elementary sub-equations:
Let’s check the two roots we found by direct substitution:
1(1−1)⋅(1−2)=00⋅(−1)=00=0\boxed{\textbf{1}} \quad (\brand{1} - 1) \cdot (\brand{1} - 2) = 0 \\ \red{0} \cdot (-1) = 0 \\ 0 = 0
2(2−1)⋅(2−2)=01⋅0=00=0 \boxed{\textbf{2}} \quad (\brand{2} - 1) \cdot (\brand{2} - 2) = 0 \\ 1 \cdot \red{0} = 0 \\ 0 = 0
We got true equalities, which proves the roots are correct. As you can see, we do not care at all what the other factors turn into, because that “whatever” still gets multiplied by zero and ends up as zero.
Let’s solve the second equation the same way:
Check the roots by substitution:
12⋅0⋅(10+0)⋅(2⋅0−1)=00⋅10⋅(−1)=00=0\boxed{\textbf{1}} \quad 2 \cdot \brand{0} \cdot (10 + \brand{0}) \cdot (2 \cdot \brand{0} - 1) = 0 \\ \red{0} \cdot 10 \cdot (-1) = 0 \\ 0 = 0
22⋅(−10)⋅(10+−10)⋅(2⋅(−10)−1)=0−20⋅0⋅(−21)=00=0 \boxed{\textbf{2}} \quad 2 \cdot (\brand{-10}) \cdot (10 + \brand{-10}) \cdot (2 \cdot (\brand{-10}) - 1) = 0 \\ -20 \cdot \red{0} \cdot (-21) = 0 \\ 0 = 0
32⋅12⋅(10+12)⋅(2⋅12−1)=01⋅212⋅0=00=0 \boxed{\textbf{3}} \quad 2 \cdot \brand{\frac{1}{2}} \cdot \left(10 + \brand{\frac{1}{2}}\right) \cdot \left(2 \cdot \brand{\frac{1}{2}} - 1\right) = 0 \\ 1 \cdot \frac{21}{2} \cdot \red{0} = 0 \\ 0 = 0
Now we’ve confirmed that all three numbers really are roots of the equation.

Zero product property

As you can see, even complicated equations can be cracked if you use the right mathematical properties. This method works beautifully and gets used all the time, both in school math and in higher math. It doesn’t really have one universally fixed name. So instead of describing the same trick over and over, let’s call it the “Zero product property”:

Zero Product Property

Any equation made of factors (brackets) whose product is zero can be solved by setting each factor equal to zero separately.

Practicing zero product property

😀
Elementary
Solve the equation:
(j−638)(j+200)=0(j-638)(j+200) = 0
Zero Product Property