# Zero Product Property

The **zero product property** states that if \(a \times b = 0 \), then \(a=0\) or \(b=0\), or both.

When factoring expressions both sides, one must be careful with cancelling the zero (null) solutions.

This can be extended to functions, and is a property to solve equations through factorization. For example, we have

\[x^2-6x+5 = 0 \quad \text{ or }\quad (x-1)(x-5) = 0.\]

Using the zero product property, either \((x-1)=0 \text{ or } (x-5) = 0.\) Thus, \(x=1\) or \(x=5.\)

However, this may not be applied to matrices as two matrices A and B can have a product of 0.

## Solve \(2(x-2)= 5x(x-2)\).

The above equation can be rewritten as follows:

\[\begin{align} 2(x-2)&= 5x(x-2)\\ 5x(x-2)-2(x-2)&=0\\ (5x-2)(x-2) &= 0 . \end{align}\]

Using the zero product property, either \((5x-2)=0\) or \((x-2) = 0.\)

Thus, \(x=\frac{2}{5}\) or \(x=2.\) \( _\square\)

## Solve \(4x^2 = 64x\).

We have

\[\begin{align} 4x^2 &= 64x \\ 4x^2 - 64x &= 0 \\ 4(x^2 - 16x) &= 0 \\ 4x(x-16) &= 0. \end{align}\]

Using the zero product property, either \(x=0\) or \((x-16) = 0.\)

Thus, \(x=0\) or \(x=16.\) \( _\square\)

## Solve \((x-2)^2(x-1) = 2(2x-5)(x-2)\).

The above equation can be rewritten as follows:

\[\begin{align} (x-2)^2(x-1) &= 2(2x-5)(x-2) \\ x^3 -5x^2 +8x -4 &= 4x^2 - 18x +20 \\ (x^3 -5x^2 +8x -4) - (4x^2 - 18x +20) &= 0 \\ x^3 - 9x^2 +26x -24 &= 0 \\ (x-2)(x-3)(x-4) &= 0. \end{align}\]

Using the zero product property, either \((x-2)=0\) or \((x-3)=0\) or \((x-4) = 0.\)

Thus, \(x=2\) or \(x=3\) or \(x=4.\) \( _\square\)

**Cite as:**Zero Product Property.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/zero-product-property/