How to solve equations by using the zero product property

Key concept: These are the classic factor-and-solve problems. Once the equation is written as a product equal to zero, each factor can be set to zero on its own. The hard part is usually just not rushing the arithmetic.

Step by step

Use the zero product property:

If $ab=0$, then $a=0$ or $b=0$.

20. $(x-5)(x+2)=0$

Set each factor to zero:

• $x-5=0 \Rightarrow x=5$
• $x+2=0 \Rightarrow x=-2$

Solutions: $x=5,-2$

21. $(2x-5)(7x+2)=0$

• $2x-5=0 \Rightarrow x=\frac52$
• $7x+2=0 \Rightarrow x=-\frac27$

Solutions: $x=\frac52,-\frac27$

22. $3(x+2)(x-2)=0$

The constant $3$ is never zero, so focus on the factors:

• $x+2=0 \Rightarrow x=-2$
• $x-2=0 \Rightarrow x=2$

Solutions: $x=-2,2$

23. $(3x-8)^2=0$

A square is zero only when its base is zero:

$3x-8=0 \Rightarrow x=\frac83.$

Solution: $x=\frac83$

Quick check

You can always test a solution by substituting it back into the original equation. If one factor becomes zero, the whole product becomes zero.

Pitfall alert

A frequent mistake is to divide both sides by a factor like $(x-5)$. That can erase a valid solution. Another one: when a factor is squared, students sometimes write two different answers, but repeated roots still give just one value unless the problem asks for multiplicity.

Try different conditions

If the equation has a coefficient in front, such as $5(x-1)(x+4)=0$, the coefficient does not affect the solutions as long as it is not zero. If the equation is not yet factored, you must factor first before using the zero product property.

Further reading

zero product property, factoring, roots