Solving Factored Quadratics with the Zero-Product Property

Welcome back. In the previous lesson, you learned to factor monic quadratic expressions by finding two integers with the required sum and product. For example,

$$x^2-3x-10=(x+2)(x-5).$$

Factoring turns one quadratic expression into a multiplication of simpler linear factors. This lesson takes the crucial next step: when that product equals zero, you can solve for the input values that make the equation true.

By the end, you will be able to solve equations such as

$$(x-7)(x+2)=0$$

by using the zero-product property. This is an algebraic method that will later help identify the zeros, or $x$-intercepts, of parabolas.

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Why a product equal to zero is special

The zero-product property states:

$$\text{If }AB=0,\text{ then }A=0\text{ or }B=0.$$

In words: if two factors multiply to make zero, at least one of the factors must itself be zero.

For ordinary numbers, this is familiar:

$$8\cdot 0=0,0\cdot (-13)=0.$$

But two nonzero real numbers cannot multiply to zero. For example,

$$8\cdot 3=24,(-2)\cdot 5=-10.$$

The same principle applies when the factors contain $x$. If

$$(x-7)(x+2)=0,$$

then either $x-7$ must be zero, or $x+2$ must be zero. Those are the only ways the entire product can be zero.

A condition matters here: the right side must be zero. You may use the zero-product property for

$$(x-7)(x+2)=0,$$

but not directly for

$$(x-7)(x+2)=12.$$

When the product is $12$, neither factor is required to be zero. The method only becomes available after the equation has been arranged with zero on one side.

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The solving routine

When an equation is already in factored form and equal to zero, follow this routine:

1. Identify the separate factors.
2. Set each factor equal to zero.
3. Solve each resulting linear equation.
4. State both solutions, unless the factors produce the same value.
5. Check by substituting if you want to confirm your result.

Consider:

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

The factors are $x-7$ and $x+2$. Apply the zero-product property:

$$x-7=0$$

or

$$x+2=0.$$

Now solve each linear equation:

$$x=7$$

or

$$x=-2.$$

Therefore,

$$\boxed{x=7\text{ or }x=-2}$$

are the two solutions.

Notice the sign reversal that happens while solving:

• $x-7=0$ gives $x=7$.
• $x+2=0$ gives $x=-2$.

The number written inside a factor is not automatically the solution. Always solve the individual linear equation.

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Factors with coefficients

The method does not change when the factors are more complicated. Each factor is still set to zero, and then you solve the resulting linear equations using ordinary inverse operations.

Solve:

$$(2x-1)(x+4)=0.$$

Set the first factor equal to zero:

$$2x-1=0.$$

Add $1$ to both sides:

$$2x=1.$$

Divide by $2$:

$$x=\frac{1}{2}.$$

Now set the other factor equal to zero:

$$x+4=0.$$

Subtract $4$:

$$x=-4.$$

So the solutions are

$$\boxed{x=\frac{1}{2}\text{ or }x=-4}.$$

A quick check shows why both work. If $x=\frac{1}{2}$, the first factor becomes zero:

$$2\left(\frac{1}{2}\right)-1=0.$$

If $x=-4$, the second factor becomes zero:

$$-4+4=0.$$

In either case, one factor is zero, so the product is zero.

A common mistake: canceling a factor

Do not divide both sides of a factored equation by one of its variable factors. For example, starting with

$$(x-7)(x+2)=0,$$

dividing by $x-7$ would appear to leave

$$x+2=0.$$

That finds $x=-2$, but it wrongly loses the other solution $x=7$. At $x=7$, the factor you divided by would equal zero, so that division was not valid. Setting each factor to zero protects both solutions.

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One solution versus two solutions

Many factored quadratic equations have two distinct solutions, because their two factors become zero at different inputs. Sometimes, however, the factors are identical.

Consider:

$$(x+6{)}^2=0.$$

This means

$$(x+6)(x+6)=0.$$

Both factors lead to the same equation:

$$x+6=0.$$

Thus,

$$\boxed{x=-6}.$$

There is only one distinct solution, even though the factor is written twice. You may hear this called a repeated solution or a double root later; for now, the important point is simply not to list the same value twice.

The pattern is:

Factored equation	Distinct solutions
$(x-3)(x+5)=0$	$x=3,x=-5$
$(x+6{)}^2=0$	$x=-6$
$x(x-4)=0$	$x=0,x=4$

The last row is worth noting: $x$ itself is a factor. Setting it equal to zero immediately gives $x=0$.

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From factoring to solving

You can now combine the previous lesson’s factoring skill with this one. Suppose you begin with a quadratic in standard form:

$$x^2-3x-10=0.$$

From the previous lesson, factor the left side. The two needed numbers multiply to $-10$ and add to $-3$, so they are $2$ and $-5$:

$$x^2-3x-10=(x+2)(x-5).$$

The equation becomes

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

Now apply the zero-product property:

$$x+2=0$$

or

$$x-5=0.$$

Therefore,

$$\boxed{x=-2\text{ or }x=5}.$$

The roles of the two skills are different:

• Factoring rewrites a quadratic as a product.
• The zero-product property turns that product equation into separate linear equations.
• Solving the linear equations gives the values of $x$.

For now, focus on recognizing when the quadratic has already been factored. In future work, you will routinely move from standard form to factored form and then use this exact method.

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Key takeaways

The zero-product property is the central idea:

$$AB=0$$

means

$$A=0\text{ or }B=0.$$

To solve a factored quadratic equation:

1. Make sure one side of the equation is zero.
2. Set each factor equal to zero.
3. Solve both linear equations.
4. Keep every distinct solution.

For example,

$$(x-7)(x+2)=0$$

has solutions

$$x=7$$

and

$$x=-2.$$

These solutions are the inputs that make the quadratic equal zero. Later in the course, that same idea will let you recognize where a parabola meets the $x$-axis.

Next, the course begins building the visual foundation for parabolas by creating a table of values and graphing the parent function

$$y=x^2.$$