and the question changes. It is no longer “which x makes this expression zero?” but “two numbers multiply to zero; what does that say about them?” That question has a sharp answer. A product can equal 12 in endless ways: 1 \cdot 12, 2 \cdot 6, 3 \cdot 4, \displaystyle (-24)\left(-\frac{1}{2}\right), and so on. It can equal 0 in only one way: one of the numbers has to be 0.

So the single hard equation splits into two easy ones, x + 1 = 0 and x - 4 = 0, each with x appearing once. That is why factoring solves equations at all, and why, whenever the expression factors, it is the fastest method there is and the first one to try.

This lesson follows the idea all the way through: the property itself, the last step that turns a factored equation into its answers, the whole method of solving a quadratic by factoring, and the area and dimension problems that the method is most often used for.

Definition

In plain words: if the product of two numbers is zero, then at least one of the numbers is zero.

In symbols, for any real numbers a and b:

\text{if } a \cdot b = 0, \text{ then } a = 0 \text{ or } b = 0.

The “or” is the inclusive one. It allows both to be zero, as in 0 \cdot 0 = 0; what it rules out is neither being zero.

The reverse direction is true as well, and simpler: a product with a factor of zero is zero, whatever the other factor is, because multiplying by zero gives zero. The zero product property is the half that goes the other way, from the product back to the factors.

Intuition

Figure 1 draws the two factors of (x + 1)(x - 4) as the dashed lines y = x + 1 and y = x - 4, and their product as the solid curve. At every x, the height of the curve is the height of one line times the height of the other.

Figure 1The product is zero exactly where one of the factors is zero, at x = -1 and x = 4.

The line y = x + 1 reaches zero at x = -1 and the line y = x - 4 at x = 4. At each of those points one factor is zero, so the product is zero, and the curve crosses the axis. Everywhere else both lines are off the axis, both heights are non-zero, and a product of two non-zero numbers is never zero, so the curve stays off the axis too. Left of -1 both factors are negative and the product is positive; between -1 and 4 they have opposite signs and the product is negative; right of 4 both are positive and so is the product. The curve can only change sign where a factor does.

Results

Zero Product Property. For any real numbers a and b,

a \cdot b = 0 \iff a = 0 \text{ or } b = 0

It extends to any number of factors. A product is zero exactly when at least one factor is zero:

\begin{aligned} a_1 a_2 \cdots a_n = 0 \iff{} & a_1 = 0 \text{ or } a_2 = 0 \\ & \text{or } \cdots \text{ or } a_n = 0 \end{aligned}

For equations this means: once one side is a product and the other side is 0, the solutions are exactly the solutions of the separate equations “factor = 0”, all of them and no others.

Solving an already-factored quadratic

Most of the work in solving a quadratic equation goes into getting it to look like this:

(x + 2)(x - 3) = 0

a product of factors on one side and a zero on the other. Try to solve it the way you would solve a linear equation and you get stuck at once: x appears in both brackets, so there is no single operation that isolates it. The zero product property finishes it instead. Reach for this step whenever an equation has a product on one side and zero on the other, whether it is a quadratic, a cubic or something larger.

Check that one side is zero. The property is a statement about the number 0 and no other. If the right side were 6 instead, the move below would not be allowed, and the equation would have to be expanded and rearranged first.

Set each factor equal to zero. The product is zero, so at least one factor is zero. Every solution must therefore satisfy

x + 2 = 0 \qquad\text{or}\qquad x - 3 = 0

The word between them is “or”. A solution only needs one factor to vanish; it does not need both, and here no single x could make both zero.

Solve each linear equation. Each is one of the easy linear equations. Subtract 2 from the first, add 3 to the second:

x = -2 \qquad\text{or}\qquad x = 3

Notice the signs. The factor x + 2 shows a plus and gives x = -2, because x = -2 is the number that makes x + 2 zero. The solution is the number that cancels what is written, not the number that is written.

Check each one in the original equation. At x = -2 the first factor is 0, and 0 \cdot (-5) = 0. At x = 3 the second factor is 0, and 5 \cdot 0 = 0. Each solution works by zeroing just one factor.

Why are these the only solutions? Because the reasoning ran in both directions. Any solution must zero a factor, so there are no others to find; and any number that zeroes a factor makes the whole product zero, so both numbers found really are solutions. On the graph of y = (x + 2)(x - 3), those are exactly the two places where the curve meets the x-axis, just as the curve in Figure 1 meets it exactly where one of its factors is zero.

Solving quadratics by factoring

A quadratic rarely arrives factored. It arrives as something like x^2 + 6 = 5x, a sum of terms spread over both sides, and the property says nothing about sums. The method is to turn it into a product equal to zero, and each step is allowed for a stated reason.

Move every term to one side, so the other side is 0. Subtracting 5x from both sides gives

x^2 - 5x + 6 = 0

Adding or subtracting the same thing on both sides never changes which numbers make an equation true, so this new equation has exactly the solutions of the old one. The zero is not decoration: it is the only number the property can use.

Factor the side that is not zero. Two numbers that multiply to 6 and add to -5 are -2 and -3, so

(x - 2)(x - 3) = 0

Factoring rewrites the same expression in a different shape; for every x the two sides of x^2 - 5x + 6 = (x - 2)(x - 3) have the same value, so again no solution is gained or lost. How to find the factors is the business of factoring trinomials, with factoring by grouping for a leading coefficient that is not 1. A common factor, including a lone x, is taken out first.

Set each factor to zero and solve. This is the step above:

x - 2 = 0 \qquad\text{or}\qquad x - 3 = 0

\text{so}\qquad x = 2 \text{ or } x = 3

Check in the original equation, not the factored one, since a factoring slip would survive a check in its own output. 2^2 + 6 = 10 = 5 \cdot 2 and 3^2 + 6 = 15 = 5 \cdot 3.

What the answers are. The numbers 2 and 3 are the solutions (or roots) of the equation x^2 - 5x + 6 = 0, the zeros of the function f(x) = x^2 - 5x + 6, and the x-intercepts of its graph, the places where y = 0. These are three names for the same numbers, and factoring finds all three at once.

The method works whenever the expression factors with integer coefficients, and then nothing is quicker. When it does not, as for x^2 + 4x + 1, the same equation is solved by completing the square or by the quadratic formula.

Area and dimension problems

A garden covers 60 square meters, and it is 4 meters longer than it is wide. How wide is it?

Two lengths are unknown, but they are tied together: once you know the width, you know the length. So one letter can name both. And the fact you are given, the area, is a product of the two lengths. A letter times an expression in that same letter makes a square term, so the equation that comes out is a quadratic. A size you know (an area, a volume, a diagonal) and a rule linking the sides turn into one quadratic equation, and the method above solves it.

Sketch the shape and name one length with a letter. Pick the length the others are described from. “The length is 4 more than the width” describes the length from the width, so let the width be w. Then the length is w + 4, and the sketch carries both.

Figure 2One letter names both sides. The area turns that into an equation.

Write the area as a product and set it equal to the given area. The area of a rectangle is length times width, as in Area of rectangles and squares:

w(w + 4) = 60

Move everything to one side. Expand and subtract 60:

w^2 + 4w - 60 = 0

The right side must be 0, not 60. A product equal to 0 forces one factor to be 0. A product equal to 60 forces nothing: 6 \cdot 10, 5 \cdot 12 and 2 \cdot 30 all make 60.

Solve the quadratic. Two numbers that multiply to -60 and add to 4 are 10 and -6:

(w + 10)(w - 6) = 0

So w = -10 or w = 6.

Keep only the roots that make every length positive. The equation does not know that w is a width. It gives every number that makes w(w + 4) equal 60, and -10 does, since (-10)(-6) = 60. But no garden is -10 meters wide, so that root is thrown out. It is not an error; it answers a question about numbers that the garden never asked.

Find every side, and check against the words, not the equation. The width is 6 m and the length is 6 + 4 = 10 m. Is the length 4 more than the width? Yes. Is the area 6 \cdot 10 = 60? Yes. Checking in the original words also catches a mistake made while setting up the equation, which checking in the equation cannot.

Why it is true

Suppose a \cdot b = 0. Either a = 0, and there is nothing more to show, or a \neq 0.

If a \neq 0, then a has a reciprocal, \displaystyle \frac{1}{a}. Multiply both sides of ab = 0 by it:

\frac{1}{a}(ab) = \frac{1}{a} \cdot 0

On the left, regroup the product: \displaystyle \frac{1}{a}(ab) = \left(\frac{1}{a} \cdot a\right) b = 1 \cdot b = b. On the right, anything times zero is zero. So

b = 0

Either way, a = 0 or b = 0. The reverse direction, that a = 0 or b = 0 makes ab = 0, is just the fact that multiplying by zero gives zero.

For three factors, group two of them: abc = (ab)c. If this is 0, the two-factor property says ab = 0 or c = 0, and if ab = 0 it says again that a = 0 or b = 0. The same grouping handles any number of factors.

Notice what the proof leans on: every non-zero number has a reciprocal. That is also why zero has none, and it is the one fact whose absence breaks the property, as the Boundaries below show.

Why solving by factoring finds every solution and no others. Each step of the method replaces an equation by one with exactly the same solutions: adding or subtracting the same thing on both sides, multiplying or dividing both sides by the same non-zero number, and rewriting an expression as an equal product, change nothing. The last step uses both directions of the property. Every solution of (x - r)(x - s) = 0 makes some factor zero, so it is r or s; and r and s each make a factor zero, so each makes the product zero. Dividing both sides by an expression containing the variable is the one move that is not safe, because it is only allowed where that expression is non-zero, and the solutions that make it zero are lost (Example 11).

Worked examples

Example 1

Two factors, already written

Solve (x + 1)(x - 4) = 0.

Solution

The left side is a product of the two numbers x + 1 and x - 4, and it equals zero, so at least one of them is zero:

x + 1 = 0 \qquad\text{or}\qquad x - 4 = 0

Each is a linear equation with x appearing once, so x = -1 or x = 4.

Check. (-1 + 1)(-1 - 4) = 0 \cdot (-5) = 0 and (4 + 1)(4 - 4) = 5 \cdot 0 = 0. Each solution makes one factor zero and leaves the other alone, which is all the product needs. These are the two points where the curve in Figure 1 crosses the axis.

Example 2

Factors with a coefficient on the variable

Solve (2x - 5)(3x + 4) = 0.

Solution

The product is zero, so one factor is zero:

\begin{aligned} 2x - 5 &= 0 & \qquad 3x + 4 &= 0 \\ 2x &= 5 & 3x &= -4 \\ x &= \frac{5}{2} & x &= -\frac{4}{3} \end{aligned}

So \displaystyle x = \frac{5}{2} or \displaystyle x = -\frac{4}{3}. The property only says a factor is zero; finding which x makes it zero is an ordinary linear equation, and its answer need not be a whole number.

Check. At \displaystyle x = \frac{5}{2}, 2x - 5 = 5 - 5 = 0. At \displaystyle x = -\frac{4}{3}, 3x + 4 = -4 + 4 = 0.

Example 3

A factor that is just the variable

Solve 4p(p + 9) = 0.

Solution

There are three factors here: 4, p and p + 9. The property says one of them is zero.

The factor 4 is never zero, so it contributes nothing. The factor p is zero when p = 0, and p + 9 is zero when p = -9. So

p = 0 \qquad\text{or}\qquad p = -9

The solution p = 0 is as real as the other one: 4 \cdot 0 \cdot (0 + 9) = 0. A factor that is the bare variable gives the solution 0, and it is the one most easily overlooked. Dividing both sides by the 4 would be harmless, because 4 is never zero; dividing by p would lose the solution p = 0, as Example 11 shows.

Example 4

A repeated factor

Solve (3a - 2)^2 = 0.

Solution

A square is a product: (3a - 2)^2 = (3a - 2)(3a - 2). Setting each factor to zero gives the same equation twice,

3a - 2 = 0 \qquad\text{so}\qquad a = \frac{2}{3}

The equation has one solution, \displaystyle a = \frac{2}{3}. It is called a double root, because it comes from two factors at once. On a graph, a double root is where the curve touches the axis and turns back instead of crossing it, as y = (x - 2)^2 does at x = 2.

Figure 3Both factors of (x - 2)^2 are the same, so there is one solution, and the curve only touches the axis.

Example 5

More than two factors, one of them a constant

Solve -4x(x - 6)(2x + 1) = 0.

Solution

The zero product property is not limited to two factors: a product of any number of factors is zero exactly when at least one of them is. The factors here are -4, x, x - 6 and 2x + 1. The constant -4 is never zero, so it contributes nothing. The other three give

\begin{aligned} x &= 0 \\ \text{or}\quad x - 6 &= 0 \\ \text{or}\quad 2x + 1 &= 0 \end{aligned}

so x = 0, x = 6 or \displaystyle x = -\frac{1}{2}. Three factors containing x, three solutions. This is a cubic, not a quadratic, and the method did not care.

Example 6

A factor that is not linear

Solve each equation.

(a) (x + 3)(x^2 - 16) = 0

(b) (x - 1)(x^2 + 9) = 0

Solution

Setting each factor to zero still works; one of the resulting equations just is not linear.

(a) x + 3 = 0 gives x = -3. The other factor gives x^2 - 16 = 0, so x^2 = 16 and x = \pm 4. That is three solutions: x = -3, x = 4 and x = -4. You could also have factored x^2 - 16 as a difference of squares, (x - 4)(x + 4), and then set all three linear factors to zero; the answers are the same.

(b) x - 1 = 0 gives x = 1. The other factor gives x^2 + 9 = 0, so x^2 = -9, and no real number squares to a negative. The only real solution is x = 1. Over the complex numbers, x^2 = -9 has the solutions x = \pm 3i, so there the equation has three solutions: x = 1 and x = \pm 3i.

Example 7

Factoring first

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

Solution

As written, the left side is a sum of three terms, not a product, and the property says nothing about sums. So factor first. Two numbers with product -10 and sum -3 are -5 and 2:

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

Now the left side is a product equal to zero, so x - 5 = 0 or x + 2 = 0, giving x = 5 or x = -2.

Check in the original equation: 5^2 - 3(5) - 10 = 25 - 15 - 10 = 0 and (-2)^2 - 3(-2) - 10 = 4 + 6 - 10 = 0.

Example 8

Terms on both sides, and a leading coefficient that is not one

Solve 6x^2 + 7x = 3.

Solution

The right side is 3, not 0, so subtract 3 from both sides first:

6x^2 + 7x - 3 = 0

To factor with a leading coefficient of 6, look for two numbers that multiply to 6 \cdot (-3) = -18 and add to 7: they are 9 and -2. Split the middle term with them and group:

\begin{aligned} 6x^2 + 9x - 2x - 3 &= 0 \\ 3x(2x + 3) - 1(2x + 3) &= 0 \\ (3x - 1)(2x + 3) &= 0 \end{aligned}

Now set each factor to zero: 3x - 1 = 0 gives \displaystyle x = \frac{1}{3}, and 2x + 3 = 0 gives \displaystyle x = -\frac{3}{2}.

Check in the original equation:

\begin{aligned} 6\left(\frac{1}{3}\right)^2 + 7\left(\frac{1}{3}\right) &= \frac{2}{3} + \frac{7}{3} = 3 \\ 6\left(-\frac{3}{2}\right)^2 + 7\left(-\frac{3}{2}\right) &= \frac{27}{2} - \frac{21}{2} = 3 \end{aligned}

Example 9

A common factor first, and a negative leading coefficient

Solve -2x^2 + 8x + 24 = 0.

Solution

Every coefficient is a multiple of -2, so take -2 out first. Taking out the negative makes the squared term inside the bracket positive, which is the shape that is easiest to factor:

\begin{aligned} -2(x^2 - 4x - 12) &= 0 \\ -2(x - 6)(x + 2) &= 0 \end{aligned}

The constant -2 is never zero, so it gives no solution; the other two factors give x = 6 or x = -2. (Dividing both sides by -2 is equally good: -2 is a non-zero number, so dividing by it cannot lose a solution.)

Check. -2(36) + 48 + 24 = 0 and -2(4) - 16 + 24 = 0.

Example 10

A product that equals something other than zero

Solve (x - 2)(x + 3) = 6.

Solution

The left side is a product, but it equals 6, not 0, and 6 can be split into two factors in endless ways: 1 \cdot 6, 2 \cdot 3, (-1)(-6), \displaystyle 12 \cdot \frac{1}{2}. Nothing tells us what either factor is. So the property cannot be used yet. First get a zero on one side: expand, then subtract 6.

\begin{aligned} x^2 + x - 6 &= 6 \\ x^2 + x - 12 &= 0 \\ (x + 4)(x - 3) &= 0 \end{aligned}

Now it applies: x = -4 or x = 3.

Check. At x = -4 the factors are -6 and -1, and (-6)(-1) = 6. At x = 3 they are 1 and 6, and 1 \cdot 6 = 6. The two solutions split 6 in two different ways, which is exactly why no factor could have been pinned down in advance.

Example 11

Why not divide by x

Solve x^3 = 9x.

Solution

Dividing both sides by x looks quick and gives x^2 = 9, so x = \pm 3. But dividing by x is only allowed when x \neq 0, and whether x can be 0 is part of the question. It can: 0^3 = 9 \cdot 0. Dividing threw that solution away.

Instead, move everything to one side and factor. Take out the common factor x, then use the difference of squares:

\begin{aligned} x^3 - 9x &= 0 \\ x(x^2 - 9) &= 0 \\ x(x - 3)(x + 3) &= 0 \end{aligned}

Three factors, so three solutions: x = 0, x = 3 or x = -3. Keeping the x as a factor, rather than dividing by it, is what keeps its solution.

Example 12

Working backwards from the solutions

Find a quadratic equation with integer coefficients whose solutions are exactly \displaystyle \frac{2}{3} and -5.

Solution

Run the property in reverse. A factor that is zero at \displaystyle x = \frac{2}{3} is \displaystyle x - \frac{2}{3}, or, multiplying by 3 to clear the fraction, 3x - 2. A factor that is zero at x = -5 is x + 5. So take

(3x - 2)(x + 5) = 0

and expand:

\begin{aligned} 3x^2 + 15x - 2x - 10 &= 0 \\ 3x^2 + 13x - 10 &= 0 \end{aligned}

Both directions of the property are at work. Multiplying by zero gives zero, so \displaystyle \frac{2}{3} and -5 are solutions. And the zero product property says any solution must make 3x - 2 or x + 5 zero, so there are no other solutions. This is how a polynomial is built from its zeros.

Example 13

A question the factored form answers

A ball is thrown upward from the edge of a roof. Its height above the ground, in meters, t seconds after it is thrown, is

h = -5(t - 4)(t + 1)

How high is the roof, and when does the ball hit the ground?

Solution

The roof is the height at the moment of the throw, t = 0:

h = -5(0 - 4)(0 + 1) = -5 \cdot (-4) \cdot 1 = 20

so the roof is 20 meters high.

The ball hits the ground when its height is zero, so solve -5(t - 4)(t + 1) = 0. The -5 is never zero, so

t - 4 = 0 \qquad\text{or}\qquad t + 1 = 0

giving t = 4 or t = -1. Both are correct solutions of the equation, but only one answers the question: t = -1 is a second before the ball was thrown, when the formula no longer describes it. The ball lands after 4 seconds.

The factored form made both questions easy. It hands you the zeros for free, which is exactly why factored form is worth having.

Example 14

A perimeter and an area

A rectangle has perimeter 34 cm and area 60 square centimeters. Find its sides.

Solution

The perimeter is twice the length plus twice the width, so the length and width add to half of 34, which is 17. If the width is w, the length is 17 - w. The area gives

\begin{aligned} w(17 - w) &= 60 \\ 17w - w^2 &= 60 \\ w^2 - 17w + 60 &= 0 \\ (w - 5)(w - 12) &= 0 \end{aligned}

So w = 5 or w = 12. This time both roots are positive, and neither is thrown out. If w = 5, the length is 12; if w = 12, the length is 5. They are the same rectangle, turned round. The sides are 5 cm and 12 cm.

Check. 2(5) + 2(12) = 34 and 5 \cdot 12 = 60.

Example 15

A table top, with a leading coefficient that is not one

The length of a rectangular table top is 3 feet less than twice its width, and its area is 90 square feet. Find its length and width.

Solution

Let the width be w. Then the length is 2w - 3, and

\begin{aligned} w(2w - 3) &= 90 \\ 2w^2 - 3w - 90 &= 0 \end{aligned}

For 2w^2 - 3w - 90, look for two numbers that multiply to 2 \cdot (-90) = -180 and add to -3: they are -15 and 12. That gives

(2w - 15)(w + 6) = 0

so \displaystyle w = \frac{15}{2} or w = -6. A width cannot be negative, so w = 7.5 feet, and the length is 2(7.5) - 3 = 12 feet.

A fractional answer is fine here. Lengths are not whole numbers just because the area is.

Check. 7.5 \cdot 12 = 90.

Example 16

A triangle

A triangular sail has an area of 28 square meters. Its base is 2 meters more than three times its height. Find the base and the height.

Solution

The base is described from the height, so let the height be h and the base 3h + 2.

Figure 4The height is h, and the base is written in terms of it.

The area of a triangle is half the base times the height, as in Area of triangles:

\frac{1}{2}(3h + 2)h = 28

Multiply both sides by 2 to clear the half, which keeps the equation true, then collect terms:

\begin{aligned} (3h + 2)h &= 56 \\ 3h^2 + 2h - 56 &= 0 \\ (3h + 14)(h - 4) &= 0 \end{aligned}

So h = 4 or \displaystyle h = -\frac{14}{3}. Only h = 4 is a height. The height is 4 m and the base is 3(4) + 2 = 14 m.

Check. \displaystyle \frac{1}{2} \cdot 14 \cdot 4 = 28.

Example 17

A frame of even width

A photo 8 inches by 12 inches sits in a frame that is the same width all the way round. The framed picture covers 192 square inches. How wide is the frame?

Solution

Call the frame’s width x. The frame adds x at each end of a side, so the outside is 12 + 2x by 8 + 2x, not 12 + x by 8 + x. The sketch is what makes that plain.

Figure 5The frame adds x on both ends of each side, so each outer side grows by 2x.

The outer area is 192:

\begin{aligned} (12 + 2x)(8 + 2x) &= 192 \\ 4x^2 + 40x + 96 &= 192 \\ 4x^2 + 40x - 96 &= 0 \\ x^2 + 10x - 24 &= 0 \end{aligned}

Dividing by 4 is allowed because it divides both sides by a non-zero number, and it makes the factoring easier:

(x + 12)(x - 2) = 0

So x = 2 or x = -12. The frame is 2 inches wide.

Check. The outside is 16 by 12, and 16 \cdot 12 = 192.

Example 18

When the answer is not a whole number

A rectangle’s length is 6 cm more than its width, and its area is 50 square centimeters. Find its sides, to the nearest tenth of a centimeter.

Solution

With width w and length w + 6:

\begin{aligned} w(w + 6) &= 50 \\ w^2 + 6w - 50 &= 0 \end{aligned}

No two integers multiply to -50 and add to 6, so this does not factor. Complete the square instead: move the 50 back across and add 9, the square of half of 6, to both sides, as in completing the square:

\begin{aligned} w^2 + 6w + 9 &= 59 \\ (w + 3)^2 &= 59 \\ w &= -3 \pm \sqrt{59} \end{aligned}

The minus sign gives a negative width, so \displaystyle w = {-3 + \sqrt{59}} \approx 4.68. The width is about 4.7 cm and the length about 10.7 cm.

Round only at the end. The exact sides, \displaystyle {-3 + \sqrt{59}} and \displaystyle {3 + \sqrt{59}}, multiply to exactly 59 - 9 = 50; the rounded ones, 4.7 \cdot 10.7 = 50.29, do not, and should not be expected to.

Example 19

A volume instead of an area

An open box is made from a square sheet of card by cutting a 3 cm square from each corner and folding up the sides. The box holds 300 cubic centimeters. How big was the sheet?

Solution

Let the sheet’s side be s. Each cut takes 3 from both ends of a side, so the floor of the box is s - 6 on each side, and the flaps fold up to make it 3 deep.

Figure 6Cut the corners, fold up the flaps. The floor is (s - 6) by (s - 6) and the box is 3 deep.

The volume of a box is floor area times height:

3(s - 6)^2 = 300

This time there is only one square term and nothing else with s in it, so there is no need to expand. Divide by 3 and take square roots, as in Solving quadratics with square roots:

\begin{aligned} (s - 6)^2 &= 100 \\ s - 6 &= \pm 10 \end{aligned}

So s = 16 or s = -4. The sheet was 16 cm square.

The condition here is stronger than “s is positive”. The floor, s - 6, must be positive too, so the sheet must be more than 6 cm across. s = 16 passes; a root such as s = 5 would not, even though it is positive.

Check. The floor is 10 by 10, and 3 \cdot 10 \cdot 10 = 300.

Boundaries

The product must equal zero. ab = 6 does not give a = 6 or b = 6, or anything else about a alone. Figure 7 shows the difference: the solutions of xy = 0 are two whole lines, each saying one coordinate is 0, while the solutions of xy = 6 form a curve on which neither coordinate is fixed.

Figure 7Every solution of xy = 0 has x = 0 or y = 0. The solutions of xy = 6 pin down neither coordinate.

It must be a product, not a sum. a + b = 0 says only that a and b are opposites. An equation like x^2 - 3x - 10 = 0 has to be factored before the property says anything.

Not every quadratic factors with integer coefficients. x^2 + 4x + 1 would have to be (x + p)(x + q) with pq = 1 and p + q = 4, but the only integers with product 1 are 1, 1 and -1, -1, and neither pair adds to 4. Its solutions, \displaystyle -2 \pm \sqrt{3}, are real, but finding them needs completing the square or the quadratic formula. Failing to factor says nothing about whether solutions exist.

A constant factor contributes no solution. In 4p(p + 9) = 0 the 4 is never zero, so only the factors containing the variable give solutions. If the constant is 0 itself, the equation says 0 = 0 and every number is a solution; that is no longer a quadratic.

Not every factor gives a real solution. A factor such as x^2 + 9 is positive for every real x and contributes nothing on the real line, although it contributes two solutions over the complex numbers.

The number of solutions can be less than the number of factors. (y - 6)^2 = 0 means (y - 6)(y - 6) = 0, and both factors say y = 6: one solution, a double root. Two different factors can also share a zero, as x - 3 and 2x - 6 do.

It needs every non-zero number to have a reciprocal. The real and complex numbers have this. Clock arithmetic does not: on a 12-hour clock, 3 \cdot 4 = 12 lands back on 0, although neither 3 nor 4 is 0. There, no number times 3 ever lands on 1, the proof has no \displaystyle \frac{1}{3} to multiply by, the property fails, and setting each factor to zero would miss solutions.

In a word problem, the root must fit the shape, not just be positive. Every length in the figure has to be positive, including the ones written as expressions: s - 6 in a folded box, 20 - 2x in a lawn with a border cut from it. A root can be positive and still make one of those negative.

Two positive roots can be one answer or two. When the two unknowns play the same part, as the length and width do in Example 14, the two roots are the same rectangle turned round. When they do not, each positive root that passes the check is a genuinely different shape, and both should be given, as in Exercise 64.

No positive root means no such shape. A rectangle with perimeter 20 has w(10 - w) = \text{area}, and w^2 - 10w + 30 = 0 has no real roots at all. So no rectangle with perimeter 20 has area 30. That is a true fact about rectangles, found by the algebra.

Rounding comes last. A rounded side, put back into the area, gives an area that is only close. Carry the exact root until the final line.

Where this goes next

The property is the last step of every factoring solution, and factoring is the first method to try on any quadratic. When the expression will not factor, the quadratic formula solves every quadratic, and choosing between the methods starts with asking whether factoring works.

Read on a graph, the property says the zeros of a polynomial are the zeros of its factors, which ties roots, zeros and x-intercepts together and is why the factored form shows the intercepts at a glance. Whether the graph crosses or only touches the axis at a zero depends on how many times its factor repeats, which is the idea of multiplicity. Run the other way, the property gives the factor theorem: c is a zero of a polynomial exactly when x - c is a factor, and it lets you write a polynomial from its zeros.

The area problems turn round into maximum and minimum problems: instead of “which width gives area 60?”, “which width gives the most area?”. There the answer is the vertex of the parabola, not a root, and projectile problems use both, a root for when a ball lands and the vertex for how high it goes. The right triangle gives another source of quadratics: a diagonal or a hypotenuse, through the Pythagorean theorem, squares both unknown sides at once. And the folded box, with the cut size as the unknown instead of the sheet, has volume x(16 - 2x)^2, a cubic. That is where modeling with polynomials starts.

The same move solves far more than quadratics. Equations like 2\sin^2\theta = \sin\theta or (\sin\theta - 1)(2\cos\theta + 1) = 0 are solved by factoring for the same reason, with a trigonometric equation in place of each linear one, and the sign changes seen in Figure 1 are how quadratic inequalities are solved.

Exercises

Factors already written

Solve.

(x - 7)(x + 2) = 0
(y + 9)(y + 1) = 0
(a - 12)(a - 5) = 0
(4n - 3)(n + 5) = 0
(2y + 7)(5y - 2) = 0
(4 - x)(x + 6) = 0

A lone variable or a constant factor

Solve.

6k(k - 8) = 0
-3w(2w + 9) = 0
-3(x + 10)(x - 1) = 0

Repeated factors and more factors

Solve.

(t + 4)^2 = 0
(2x + 3)^2 = 0
(x - 3)(2x - 6) = 0. How many solutions are there?
(x - 1)(x + 6)(2x - 3) = 0
2z(z + 3)(z - 7) = 0
(y + 5)(y^2 - 49) = 0
(x^2 + 4)(x - 3) = 0. Give the real solutions, then any complex ones.

Factor first

Solve.

x^2 + 9x + 20 = 0
x^2 - 2x - 24 = 0
p^2 - 11p = 0
m^2 = 5m + 36
y^2 + 49 = 14y
2x^2 - 5x - 3 = 0
10x^2 + x = 2
3x^2 + 12x - 36 = 0
4x^2 = 9
x(x + 2) = 2(x + 8)
(x + 1)(x - 5) = 7
x^3 - 16x = 0
n^3 + 3n^2 - 10n = 0

Using the property

The product of two consecutive integers is 72. Find the integers.
Where does the graph of y = x^2 - x - 12 cross the x-axis?
Where does the graph of y = (x + 4)(x - 1) meet the x-axis?
Write a quadratic equation with integer coefficients whose solutions are exactly 4 and \displaystyle -\frac{1}{2}.
Solve x^2 = 4x.
A stall’s weekly profit, in dollars, from selling n scarves is P = -2(n - 10)(n - 40). For which numbers of scarves does the stall exactly break even, with a profit of zero?

Rectangles

A rectangle is 5 cm longer than it is wide, and its area is 84 square centimeters. Find its width and length.
A rectangular rug is 4 feet narrower than it is long, and covers 96 square feet. Find its dimensions.
A rectangle’s length is twice its width, and its area is 72 square centimeters. Find its sides.
A rectangle’s length is 1 more than twice its width, and its area is 105. Find its length and width.
A rectangle has perimeter 30 m and area 56 square meters. Find its sides.
Each side of a square is increased by 3 cm, and the new square has area 121 square centimeters. How long was each side of the original square?

Triangles and diagonals

A triangle’s height is 3 cm more than its base, and its area is 20 square centimeters. Find the base and the height.
A triangle’s base is twice its height, and its area is 49 square inches. Find the base and the height.
One leg of a right triangle is 7 cm longer than the other, and the triangle’s area is 30 square centimeters. Find both legs.
A rectangle’s length is 17 m more than its width, and its diagonal is 25 m. Find its length and width.

Borders, paths and boxes

A garden 10 m by 6 m has a path of the same width all the way round its outside. Garden and path together cover 96 square meters. How wide is the path?
A lawn 20 m by 15 m has a flower border of the same width cut along all four of its edges, on the inside. The grass left in the middle covers 204 square meters. How wide is the border?
An open box is made from a square sheet by cutting a 4 cm square from each corner and folding up the sides. The box holds 256 cubic centimeters. How big was the sheet?
An open box is made from a sheet 20 cm by 14 cm by cutting a square of side x from each corner. The floor of the box has area 160 square centimeters. Find x and the volume of the box.

Answers that are not whole numbers

Give each length to the nearest tenth.

A rectangle’s length is 3 m more than its width, and its area is 20 square meters. Find the width and the length.
A triangle’s base is 4 feet more than its height, and its area is 40 square feet. Find the height and the base.
A rectangle’s length is twice its width, and its diagonal is 10 inches. Find the width and the length.

Reasoning

Explain why (x - 2)(x + 3) = 6 cannot be solved by setting x - 2 = 6 or x + 3 = 6.
The property says “a = 0 or b = 0”, not “a = 0 and b = 0”. Explain why “and” would be wrong, using x(x - 5) = 0.
Without multiplying anything out, how many real solutions does (x^2 + 1)(x - 2)^2 = 0 have? Explain.
Someone claims that (x - a)(x - b) = 0 always has two different solutions. When is the claim false, and what happens then?
Explain why the 3 in 3(x - 2)(x + 8) = 0 adds no solution, but the x in x(x - 2)(x + 8) = 0 does.
Explain why x = r and x = s are solutions of (x - r)(x - s) = 0, and why no other number is.
Using only the two-factor property, show that if abc = 0 then a = 0, b = 0 or c = 0.
On a 12-hour clock, 3 \cdot 4 lands on 0. Which step of the proof in “Why it is true” fails there, and why?
Explain why x^2 + 4x + 1 = 0 cannot be solved by factoring with integers, then solve it another way.
In Exercise 47 both roots of the equation are positive, yet one is thrown out. Explain which one, and why being positive was not enough.
Show that if a rectangle’s length is k more than its width, where k > 0, then for any positive area A the equation w(w + k) = A has exactly one positive root.
A rectangular pen uses a long wall as one side and 26 m of fence for the other three, and it encloses 60 square meters. Find every pen that works, and explain why this time both roots are kept as different answers.
A 12 by 8 photo is given a frame of the same width all round, and the framed picture is said to cover 90 square inches. Without solving, explain why this is impossible.

Answers

x = 7 or x = -2. One factor is zero: x - 7 = 0 or x + 2 = 0.
y = -9 or y = -1. Set y + 9 = 0 and y + 1 = 0.
a = 12 or a = 5. Set a - 12 = 0 and a - 5 = 0.
\displaystyle n = \frac{3}{4} or n = -5. From 4n - 3 = 0, 4n = 3; from n + 5 = 0, n = -5.
\displaystyle y = -\frac{7}{2} or \displaystyle y = \frac{2}{5}. From 2y + 7 = 0, 2y = -7; from 5y - 2 = 0, 5y = 2.
x = 4 or x = -6. The factor 4 - x is zero when x = 4; the order inside the bracket does not matter.
k = 0 or k = 8. The factor 6 is never zero; k = 0 and k - 8 = 0 give the solutions.
w = 0 or \displaystyle w = -\frac{9}{2}. The factor -3 is never zero; w = 0, or 2w + 9 = 0 gives 2w = -9.
x = -10 or x = 1. The -3 contributes nothing.
t = -4, the only solution, a double root. Both factors of (t + 4)(t + 4) say t + 4 = 0.
\displaystyle x = -\frac{3}{2}, a double root. From 2x + 3 = 0, 2x = -3.
One solution, x = 3. The first factor gives x = 3, and 2x - 6 = 0 gives 2x = 6, so x = 3 again. The two factors share their zero: 2x - 6 = 2(x - 3), so the equation is 2(x - 3)^2 = 0 and x = 3 is a double root.
x = 1, x = -6 or \displaystyle x = \frac{3}{2}. One of the three factors must be zero.
z = 0, z = -3 or z = 7. The 2 contributes nothing; the other three factors each give one solution.
y = -5, y = 7 or y = -7. From y^2 - 49 = 0, y^2 = 49, so y = \pm 7.
The only real solution is x = 3. The factor x^2 + 4 is zero when x^2 = -4, which has no real solution; over the complex numbers it gives x = \pm 2i.
x = -4 or x = -5. Factor: (x + 4)(x + 5) = 0.
x = 6 or x = -4. Factor: (x - 6)(x + 4) = 0.
p = 0 or p = 11. Take out the common factor: p(p - 11) = 0.
m = 9 or m = -4. Rearrange to m^2 - 5m - 36 = 0, then factor: (m - 9)(m + 4) = 0.
y = 7, the only solution. Rearrange to y^2 - 14y + 49 = 0, which is (y - 7)^2 = 0.
\displaystyle x = -\frac{1}{2} or x = 3. Two numbers with product 2 \cdot (-3) = -6 and sum -5 are -6 and 1, which gives (2x + 1)(x - 3) = 0.
\displaystyle x = \frac{2}{5} or \displaystyle x = -\frac{1}{2}. Rearrange to 10x^2 + x - 2 = 0; two numbers with product -20 and sum 1 are 5 and -4, which gives (5x - 2)(2x + 1) = 0.
x = -6 or x = 2. Take out the common factor 3: 3(x^2 + 4x - 12) = 3(x + 6)(x - 2) = 0.
\displaystyle x = \frac{3}{2} or \displaystyle x = -\frac{3}{2}. Rearrange to 4x^2 - 9 = 0, a difference of squares: (2x - 3)(2x + 3) = 0.
x = 4 or x = -4. Expand: x^2 + 2x = 2x + 16, so x^2 - 16 = 0 and (x - 4)(x + 4) = 0.
x = 6 or x = -2. Expand: x^2 - 4x - 5 = 7, so x^2 - 4x - 12 = 0 and (x - 6)(x + 2) = 0. Check: 7 \cdot 1 = 7 and (-1)(-7) = 7.
x = 0, x = 4 or x = -4. x(x^2 - 16) = 0, so x(x - 4)(x + 4) = 0.
n = 0, n = -5 or n = 2. n(n^2 + 3n - 10) = 0, so n(n + 5)(n - 2) = 0.
8 and 9, or -9 and -8. With n and n + 1: n(n + 1) = 72, so n^2 + n - 72 = 0 and (n + 9)(n - 8) = 0, giving n = 8 or n = -9.
At (4, 0) and (-3, 0). The graph meets the axis where y = 0: x^2 - x - 12 = (x - 4)(x + 3) = 0.
At (-4, 0) and (1, 0). The graph meets the axis where y = 0, that is where (x + 4)(x - 1) = 0, so x = -4 or x = 1.
2x^2 - 7x - 4 = 0. Factors zero at 4 and \displaystyle -\frac{1}{2} are x - 4 and 2x + 1, and (x - 4)(2x + 1) = 2x^2 - 7x - 4. Any non-zero multiple of it also works.
x = 0 or x = 4. Rearrange to x^2 - 4x = 0, so x(x - 4) = 0. Dividing by x instead would lose x = 0.
n = 10 or n = 40 scarves. Set P = 0; the -2 is never zero, so n - 10 = 0 or n - 40 = 0.
Width 7 cm, length 12 cm. w(w + 5) = 84, so w^2 + 5w - 84 = 0 and (w + 12)(w - 7) = 0. A width cannot be -12, so w = 7.
12 feet by 8 feet. With length l: l(l - 4) = 96, so (l - 12)(l + 8) = 0 and l = 12.
6 cm by 12 cm. w \cdot 2w = 72, so w^2 = 36 and w = 6.
Width 7, length 15. w(2w + 1) = 105, so 2w^2 + w - 105 = (2w + 15)(w - 7) = 0.
7 m and 8 m. The sides add to 15: w(15 - w) = 56, so (w - 7)(w - 8) = 0. The two roots are the same rectangle turned round.
8 cm. (s + 3)^2 = 121, so s + 3 = \pm 11, and only s = 8 is positive.
Base 5 cm, height 8 cm. \displaystyle \frac{1}{2}b(b + 3) = 20, so b^2 + 3b - 40 = (b + 8)(b - 5) = 0.
Base 14 inches, height 7 inches. \displaystyle \frac{1}{2}(2h)h = h^2 = 49, so h = 7.
5 cm and 12 cm. \displaystyle \frac{1}{2}x(x + 7) = 30, so x^2 + 7x - 60 = (x + 12)(x - 5) = 0.
7 m by 24 m. By the Pythagorean theorem w^2 + (w + 17)^2 = 625, so w^2 + 17w - 168 = (w + 24)(w - 7) = 0.
1 m. (10 + 2x)(6 + 2x) = 96 gives x^2 + 8x - 9 = (x + 9)(x - 1) = 0.
1.5 m. (20 - 2x)(15 - 2x) = 204 gives 2x^2 - 35x + 48 = (2x - 3)(x - 16) = 0, so \displaystyle x = \frac{3}{2} or x = 16; a border 16 m wide would leave 20 - 32 < 0.
16 cm square. 4(s - 8)^2 = 256, so (s - 8)^2 = 64 and s - 8 = \pm 8; s = 0 is no sheet.
x = 2 cm, and the volume is 320 cubic centimeters. (20 - 2x)(14 - 2x) = 160 gives x^2 - 17x + 30 = (x - 2)(x - 15) = 0; x = 15 would cut 30 cm from a 14 cm side. The box is 16 by 10 by 2.
Width 3.2 m, length 6.2 m. w^2 + 3w - 20 = 0, so \displaystyle w = \frac{-3 + \sqrt{89}}{2} \approx 3.22.
Height 7.2 feet, base 11.2 feet. \displaystyle \frac{1}{2}h(h + 4) = 40, so h^2 + 4h - 80 = 0 and \displaystyle h = {-2 + 2\sqrt{21}} \approx 7.17.
Width 4.5 inches, length 8.9 inches. w^2 + (2w)^2 = 100, so w^2 = 20, \displaystyle w = 2\sqrt{5} \approx 4.47 and the length is \displaystyle 4\sqrt{5} \approx 8.94.
The property is about a product equal to 0 only. A product of 6 can come from 1 \cdot 6, 2 \cdot 3, (-1)(-6) and endless other pairs, so neither factor has to be 6. Indeed x - 2 = 6 gives x = 8, and then the product is 6 \cdot 11 = 66. The real solutions, x = -4 and x = 3, come from expanding and moving the 6 across (Example 10).
At x = 5 the factors are 5 and 0: only one is zero, yet the product is 0, so x = 5 is a solution. With “and”, both x = 0 and x - 5 = 0 would have to hold at once, which no number does, and the equation would wrongly seem to have no solutions.
One, x = 2. The factor x^2 + 1 is at least 1 for every real x, so it is never zero and gives no real solution. The factor (x - 2)^2 is (x - 2)(x - 2), and both copies are zero only at x = 2.
It is false when a = b. Then the equation is (x - a)^2 = 0, both factors are zero at the same number, and there is only one solution, x = a, a double root.
The zero product property says one of the factors must be zero. The 3 is a fixed non-zero number, so it can never be that factor, and it contributes nothing. The x is zero when x = 0, which makes the whole product zero, so x = 0 is a solution.
At x = r the first factor is 0, and 0 times any number is 0; likewise at x = s. Conversely, if x is any number other than r and s, both factors x - r and x - s are non-zero, and a product of two non-zero numbers is never zero, so the equation fails.
Write abc = (ab)c. If it is 0, the two-factor property gives ab = 0 or c = 0. In the first case, the property again gives a = 0 or b = 0. So one of a, b, c is 0.
The step that multiplies by \displaystyle \frac{1}{a}. On a 12-hour clock, 3 times any number lands on 0, 3, 6 or 9, never on 1, so 3 has no reciprocal and there is nothing to multiply 3 \cdot 4 = 0 by to recover 4 = 0.
\displaystyle x = -2 \pm \sqrt{3}. Integer factors (x + p)(x + q) would need pq = 1 and p + q = 4, but the only integer pairs with product 1 are 1, 1 and -1, -1, with sums 2 and -2. Completing the square instead: x^2 + 4x + 4 = 3, so (x + 2)^2 = 3 and \displaystyle x + 2 = \pm\sqrt{3}.
x = 16 is thrown out. The border comes off both ends of each side, so the middle is 20 - 2x by 15 - 2x, and at x = 16 both of those are negative. Every length in the figure must be positive, not just x.
The equation is w^2 + kw - A = 0, so by the quadratic formula \displaystyle w = \frac{-k \pm \sqrt{k^2 + 4A}}{2}. Since A > 0, \displaystyle \sqrt{k^2 + 4A} is real and larger than k, so the + root is positive and the - root is negative.
Two pens: 3 m out from the wall and 20 m along it, or 10 m out and 6 m along. With x for each side meeting the wall, the side along it is 26 - 2x, and x(26 - 2x) = 60 gives x^2 - 13x + 30 = (x - 3)(x - 10) = 0. Both roots keep 26 - 2x positive. The two sides play different parts, one fenced twice and one fenced once, so x = 3 and x = 10 are two different pens, not one pen turned round.
The photo alone covers 12 \cdot 8 = 96 square inches, and a frame only adds to that, so the framed picture covers at least 96. It cannot cover 90.