Solving by Factoring

You can often solve an equation by factoring, using the Zero Product Rule.

The Zero Product Rule. If a product of things equals 0, then at least one of the things must equal 0.

In terms of symbols, this means that

$$\quad\hbox{if}\quad a b = 0, \quad\hbox{then}\quad a = 0 \quad\hbox{or}\quad b = 0.$$

This works with a product of more than two things.

If you can get an equation into the form "$\hbox{JUNK} = 0$ ", you can apply the Zero Product Rule by factoring JUNK and setting each factor equal to 0.


Example. Solve $x^2 - 2 x - 3 = 0$ .

$$\matrix{ & x^2 - 2 x - 3 & = & 0 & \cr & (x - 3)(x + 1) & = & 0 & \cr & \swarrow & & \searrow & \cr x - 3 = 0 & & & & x + 1 = 0 \cr x = 3 & & & & x = -1 \cr}$$

The solutions are $x = 3$ and $x = -1$ .


Example. Solve $(x - 1)(x - 2)(x - \sqrt{17}) =
   0$ .

The equation is already in "$\hbox{JUNK} = 0$ " form, and the left side is factored.

$$\matrix{ & (x - 1)(x - 2)(x - \sqrt{17}) & = & 0 & \cr & \swarrow & \downarrow & \searrow & \cr x - 1 = 0 & & x - 2 = 0 & & x - \sqrt{17} = 0 \cr x = 1 & & x = 2 & & x = \sqrt{17} \cr}$$

The solutions are $x = 1$ , $x = 2$ , and $x =
   \sqrt{17}$ .


Example. Solve $x^3 - 4 x = 0$ .

$$\matrix{ & x^3 - 4 x & = & 0 & \cr & x(x^2 - 4) & = & 0 & \cr & x(x - 2)(x + 2) & = & 0 & \cr & \swarrow & \downarrow & \searrow & \cr x = 0 & & x - 2 = 0 & & x + 2 = 0 \cr & & x = 2 & & x = -2 \cr}$$

The solutions are $x = 0$ , $x = 2$ , and $x =
   -2$ .


Example. Solve $x^2 + 4 x = -3$ .

You can't use the Zero Product Rule unless you have 0 on one side of the equation. In this case, I have to move the -3 over first.

$$\matrix{ & x^2 & + & 4 x & & & = & -3 \cr + & & & & & 3 & & 3 \cr \noalign{\vskip2pt\hrule\vskip2pt} & x^2 & + & 4 x & + & 3 & = & 0 \cr}$$

Now I can factor:

$$\matrix{ & x^2 + 4 x + 3 & = & 0 & \cr & (x + 1)(x + 3) & = & 0 & \cr & \swarrow & & \searrow & \cr x + 1 = 0 & & & & x + 3 = 0 \cr x = -1 & & & & x = -3 \cr}$$

The solutions are $x =
   -1$ and $x = -3$ .


Example. Solve $6 x^3 - 9 x^2 = 6 x$ .

$$\eqalign{6 x^3 - 9 x^2 &= 6 x \cr 6 x^3 - 9 x^2 - 6 x &= 0 \cr 3 x(2 x^2 - 3 x - 2) &= 0 \cr 3 x(2 x + 1)(x - 2) &= 0 \cr}$$

$3 x = 0$ gives $x = 0$ .

$2 x + 1 = 0$ gives $2 x = -1$ , or $x = -\dfrac{1}{2}$ .

$x - 2 = 0$ gives $x = 2$ .

The solutions are $x = 0$ , $x = -\dfrac{1}{2}$ , and $x = 2$ .


Example. Solve $(2 x + 1)(x - 3) = 5 x(x -
   3)$ .

Warning: Don't cancel the $x - 3$ 's from both sides! Instead, move everything to one side, then factor:

$$\eqalign{ (2 x + 1)(x - 3) &= 5 x(x - 3) \cr (2 x + 1)(x - 3) - 5 x(x - 3) &= 0 \cr (x - 3)[(2 x + 1) - 5 x] &= 0 \cr (x - 3)(1 - 3 x) &= 0 \cr}$$

$x - 3 = 0$ gives $x = 3$ .

$1 - 3 x = 0$ gives $3 x = 1$ , or $x = \dfrac{1}{3}$ .

The solutions are $x = 3$ and $x = \dfrac{1}{3}$ .


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