Split a quadratic into two factors, then use the fact that a product of zero means at least one factor is zero.
A ball rolls off a table and hits the floor. The height of the ball in feet at time t seconds is modeled by h(t)=t2−5t+6. A physics student asks: at what exact moments is the ball at ground level? That question becomes "when does h(t)=0?" — and answering it means solving a quadratic equation by factoring.
A quadratic equation in standard form looks like this:
ax2+bx+c=0
When the left side factors into two binomials, solving becomes straightforward. The key is a single rule called the zero product property: if two numbers multiply to give zero, then at least one of them must be zero. In symbols, if A⋅B=0, then either A=0 or B=0. There is no other way to multiply two things and get zero — you cannot get zero by multiplying two nonzero numbers together.
The strategy is to factor the quadratic into the form (x−r)(x−s)=0, and then set each factor equal to zero separately. The two solutions you get, x=r and x=s, are called the roots or zeros of the quadratic. They are the x-values where the parabola crosses the x-axis.
To factor a trinomial x2+bx+c, you need two numbers that multiply to c and add to b. Once you find that pair, you write the factors and apply the zero product property.
Solving x² - 5x + 6 = 0
x2−5x+6=0
The equation is already in standard form with zero on the right side. That is where it needs to be before you factor.
−2⋅(−3)=6and−2+(−3)=−5
You need two numbers that multiply to 6 and add to -5. Try -2 and -3. Their product is 6 and their sum is -5. That is the pair.
(x−2)(x−3)=0
Write the factored form. Each number from the pair goes into one binomial.
x−2=0orx−3=0
Zero product property: at least one factor must be zero, so set each one equal to zero.
x=2orx=3✓
Solve each mini-equation by adding to both sides. These are the two roots.
Here is the same problem expressed as a function. The zeros of f(x)=x2−5x+6 are the x-values where the graph crosses the x-axis. The graph confirms that the parabola hits zero at exactly x=2 and x=3.
Interactive graph — scroll to zoom, drag to pan
Now try a problem where the equation is not already in standard form. You must move everything to one side before you can factor.
Solving x² + 3x = 10
x2+3x−10=0
Subtract 10 from both sides to get zero on the right. You cannot factor until zero is alone on one side.
5⋅(−2)=−10and5+(−2)=3
You need two numbers that multiply to -10 and add to 3. Try 5 and -2.
(x+5)(x−2)=0
Write the factored form.
x+5=0orx−2=0
Set each factor equal to zero.
x=−5orx=2✓
Solve each equation. Two solutions, and both are valid.
Sometimes the two roots are the same number. That happens when the quadratic is a perfect square trinomial — the parabola just touches the x-axis at one point instead of crossing it at two.
Solving x² - 6x + 9 = 0
x2−6x+9=0
Already in standard form.
−3⋅(−3)=9and−3+(−3)=−6
Both numbers in the pair are -3. That tells you this is a perfect square.
(x−3)(x−3)=0
Both factors are identical.
(x−3)2=0
You can write it this way too — it makes the perfect square structure visible.
x−3=0
Even though there are two factors, they give the same equation.
x=3✓
One repeated root. The parabola touches the x-axis at exactly one point.
Interactive graph — scroll to zoom, drag to pan
Solve each equation by factoring.
Practice Questions
x2+7x+12=0
x2−x=12
x2−8x=0
Regents Corner
On Part II and Part III of the Algebra I Regents, solving a quadratic by factoring is a multi-step problem worth 2 or 3 points. Graders look for a correctly factored expression, the zero product property applied explicitly, and both solutions stated. Missing one root loses a point even if your factoring is right.
A student sees x² - 8x = 0, divides both sides by x, and gets x - 8 = 0, so x = 8. That misses x = 0 entirely. Dividing by a variable is only valid if you know for certain the variable is not zero — and here it can be zero. Always factor instead of dividing by x.