A quadratic equation is one that can be written ax² + bx + c = 0 with a ≠ 0. Every method in this course ends at the same place: a product equal to zero, where the factors can be taken one at a time. The property that licenses that step holds for zero and for no other number.
Core ideas
The manipulation in this lesson is one line long. Knowing when that line is legal is the lesson.
3x² = 5x - 2 and (x + 1)(x - 4) = 6 are both quadratic, because each can be rearranged into ax² + bx + c = 0 with a ≠ 0. The rearranging is not cosmetic: every method in this course is stated for that form and guaranteed only for it.
ab = 0 forces one of the two to be zero. ab = 12 forces nothing at all: 3 · 4, 2 · 6 and 24 · (1/2) are all 12. What is special about zero is that a product cannot arrive there unless a factor is already there.
Linear Equations and Inequalities defined a solution as a value that makes the statement true, and nothing changes. A quadratic usually has two solutions, so an answer carrying one number is usually half an answer, and the check is the same as it always was: substitute, and see.
The material
The definition, the property, its one-line proof, and the two places it gets misapplied.
A quadratic equation in x is one that can be written ax² + bx + c = 0, where a, b and c are numbers and a ≠ 0. That arrangement is called standard form.
The condition a ≠ 0 is what makes the equation quadratic rather than linear. If a were 0 the x² term would disappear and Linear Equations and Inequalities would already have solved what was left.
b or c may be zero, and often are. x² - 9 = 0 has b = 0; x² + 3x = 0 has c = 0. Both are quadratic, and each of those two shapes gets a method of its own later in the course.
For real numbers a and b, ab = 0 if and only if a = 0 or b = 0.
The "or" is inclusive: both may be zero, as in 0 · 0 = 0. The statement rules out only the case where neither is.
One direction is arithmetic. If a = 0 then ab = 0 · b = 0, and the same for b = 0. So a product with a zero factor is zero.
The other direction is the useful one. Suppose ab = 0 and a ≠ 0. Because a is a non-zero real number it has a reciprocal 1/a, so both sides may be multiplied by it: (1/a)(ab) = (1/a) · 0.
The right-hand side is 0. The left-hand side regroups, by associativity, as ((1/a) · a)b = 1 · b = b. So b = 0. Assuming a ≠ 0 forced b = 0, which is exactly the claim that a = 0 or b = 0.
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The proof leaned on one property of the real numbers that is easy to take for granted: every non-zero number has a reciprocal. The rule can fail in systems without that property. In arithmetic modulo 12 — clock arithmetic — 3 · 4 = 12, which is 0 on that clock, and neither 3 nor 4 is zero there. So the zero product property is a fact about the numbers you are using, not a fact about algebra in general.
(x - 3)(x + 5) = 9 is a perfectly good equation, and the property says nothing whatever about it. Setting x - 3 = 9 gives x = 12, and substituting back gives (12 - 3)(12 + 5) = 9 · 17 = 153, not 9. That was not an arithmetic slip. It was an appeal to a theorem whose hypothesis had not been met.
The repair is always the same: expand, move everything to one side, and produce a zero. (x - 3)(x + 5) = 9 becomes x² + 2x - 15 = 9, then x² + 2x - 24 = 0, and “Solving by Factoring” finishes it.
The property extends to any finite number of factors by being applied repeatedly: abc = 0 means a = 0 or bc = 0, and the second case splits again. So x(x - 2)(x + 7) = 0 has the three solutions 0, 2 and -7, found one bracket at a time.
x² = 5x invites you to divide both sides by x, giving x = 5. But x = 0 also satisfies the original equation, since 0² = 5 · 0. Dividing by x quietly assumed x ≠ 0, and the solution that broke the assumption was thrown away by the step that made it.
Division by a variable is legal only when you know the variable is not zero, and in an equation you are still solving you do not know that — finding out is the point. Move everything to one side instead: x² - 5x = 0, so x(x - 5) = 0, so x = 0 or x = 5. Nothing is lost, and the method is one line longer.
(2x - 1)(x + 4) = 0. The property gives 2x - 1 = 0 or x + 4 = 0, which are two linear equations of the kind Linear Equations and Inequalities solved in a line each. The first gives x = 1/2, the second x = -4.
Those are the only solutions. Any other value of x makes both brackets non-zero, and a product of two non-zero real numbers is not zero — which is the first direction of the theorem, read as a guarantee that nothing has been missed.
Interactive
Every root comes from one factor being zero
Method
Four steps, and the first two exist only to make the third legal.
Expand any brackets, then move every term to the same side. Until the other side reads 0, no factorisation you write down is of any use.
Polynomials and Factoring supplies the techniques and “Solving by Factoring” applies them. What you need is a product of polynomials; a sum, however tidy, cannot be split by this property.
One factor at a time, each giving its own small equation. A constant factor such as the 5 in 5(x - 2)(x + 3) = 0 yields nothing, because 5 = 0 is false.
Each factor is linear, so Linear Equations and Inequalities finishes it. Substitute back into the equation you were given, not the rearranged one — that is what catches an illegal rearrangement as well as an arithmetic slip.
(x - 3)(x + 5) = 9. The left-hand side arrives already factored, which is exactly what makes the wrong route attractive.
The wrong route returned x = 4, which is a genuine solution, alongside x = 12, which is not. That coincidence is worth sitting with. A method that happens to produce a correct answer has not been validated by producing it, and here only the check separates the two numbers.
Notice what the right route did with the factored form it was handed: it threw it away and expanded. That looks wasteful and is not, because the factorisation you need is the one whose product is zero, not the one you happened to be given.
For a faded zero-product pass, solve 5(x - 2)(2x + 1) = 0. The supplied decision is that the hypothesis is already met: the product equals zero. Decide which factor contributes no solution, solve the other two and check both values in the original. Compare with x = 2 and x = -1/2; if 5 appeared as a third answer, a constant factor was treated as though it contained x.
Given any equation with a factored side you should look at the right-hand side before doing anything else, and rearrange whenever it is not 0. You should also be able to give the one-line argument with the reciprocal, because that argument is what tells you the property cannot be stretched to cover ab = 12.
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