A student has the equation x² − 5x = 6. They factor the left side as x(x − 5) = 6 and conclude x = 6 or x − 5 = 6, giving x = 6 or x = 11. What error did they make?
AThey factored the left side incorrectly
BThey applied the zero product property without first setting the equation equal to zero
CThey should have used the quadratic formula for this equation
DThey found two solutions when this equation has only one
The zero product property states that if a product equals zero, one of the factors must be zero. It does NOT apply when a product equals any other number — if two things multiply to 6, there are infinitely many possibilities. The required first step is to move all terms to one side: x² − 5x − 6 = 0, then factor to (x − 6)(x + 1) = 0, giving x = 6 or x = −1. The student's error of skipping this step is the most common mistake in solving quadratics by factoring.
Question 2 Multiple Choice
Which of the following quadratic equations has exactly one real solution?
Ax² − 5x + 6 = 0
Bx² − 6x + 9 = 0
Cx² − 4 = 0
Dx² + 2x − 8 = 0
x² − 6x + 9 = 0 factors as (x − 3)² = 0, giving x = 3 as a repeated root — one solution (technically multiplicity 2, but only one distinct value). The others factor as (x−2)(x−3) = 0 (two solutions: 2 and 3), (x−2)(x+2) = 0 (two solutions: 2 and −2), and (x+4)(x−2) = 0 (two solutions: −4 and 2). A repeated factor means the parabola touches the x-axis at exactly one point without crossing it.
Question 3 True / False
To solve the equation (x + 4)(x − 3) = 0, you can set each factor equal to zero independently because the product equals zero.
TTrue
FFalse
Answer: True
This is the zero product property in action. Because the product of the two factors is zero, at least one factor must be zero — there is no other way a product can be zero. Setting x + 4 = 0 gives x = −4, and setting x − 3 = 0 gives x = 3. Both values satisfy the original equation. The reason you can treat the factors independently is that zero is the only number with this special 'forcing' property.
Question 4 True / False
If (x + 2)(x − 3) = 6, you can find the solutions by setting x + 2 = 6 or x − 3 = 6.
TTrue
FFalse
Answer: False
The zero product property only applies when a product equals zero, not any other number. If x + 2 = 6 then x = 4, and checking: (4+2)(4−3) = 6×1 = 6, so x = 4 happens to work — but x − 3 = 6 gives x = 9, and (9+2)(9−3) = 11×6 = 66 ≠ 6. The correct method is to expand and rearrange: x² − x − 6 = 6 → x² − x − 12 = 0 → (x − 4)(x + 3) = 0, giving x = 4 or x = −3.
Question 5 Short Answer
Why must a quadratic equation be set equal to zero before you can use the zero product property? What goes wrong if you skip this step?
Think about your answer, then reveal below.
Model answer: The zero product property is specifically about products that equal zero: if ab = 0, then a = 0 or b = 0. This works because zero is the only number with the property that one factor must be zero. If the product equals any other number — say 6 — then there are infinitely many factor pairs that multiply to 6, so you cannot set each factor equal to that number and solve independently. Skipping the 'set equal to zero' step leads to equations like x(x−5) = 6, which incorrectly suggests x = 6 or x−5 = 6 as solutions.
The zero product property is a uniquely powerful tool precisely because it only works for zero. Any equation of the form ab = k where k ≠ 0 cannot be solved by splitting into a = k or b = k. Setting the equation equal to zero first converts the problem into exactly the form where this property applies, turning the quadratic into two simple linear equations.