Solving Quadratic Equations by Factoring

Think about two numbers whose product is 0. If neither number were 0, their product could not be 0: $3\times 5=15$, $(-2)(0.1)=-0.2$, and no amount of searching finds two non-zero numbers that multiply to 0. So at least one of them must be 0. This is the zero product property: if $A\times B=0$, then $A=0$ or $B=0$ (or both). The property is special to zero. If you know only that $A\times B=6$, the two factors could be $2$ and $3$, or $1$ and $6$, or $-2$ and $-3$, or $12$ and $0.5$, and so on. Knowing that a product is 6 tells you nothing definite about either factor. Knowing that a product is 0 tells you a lot.

Notice the signs: the factor $(x+2)$ gives the root $x=-2$, because $-2$ is the number that makes that bracket zero. Always verify your roots by substituting each one into the original equation. For $x=-2$: ${(-2)}^2+10(-2)+16=4-20+16=0\text{.}$ For $x=-8$: ${(-8)}^2+10(-8)+16=64-80+16=0\text{.}$ Both make the left side 0, so both are roots.

Solving a quadratic equation by factoring | Algebra II | Khan Academy

Every point on the $x$-axis has $y=0$. So the points where the parabola $y=x^2+10x+16$ meets the $x$-axis are exactly the $x$-values that solve $x^2+10x+16=0$. Solving the equation and finding the zeros of the relation are the same job. This is why the factored form $y=a(x-r)(x-s)$ from the last unit shows the zeros directly: they are $x=r$ and $x=s$. Here $y=(x+2)(x+8)=(x-(-2))(x-(-8))\text{,}$ so $r=-2$ and $s=-8$. The axis of symmetry is halfway between the zeros, at $x=\frac{-2+(-8)}{2}=-5$, and the vertex is $(-5,(-3)(3))=(-5,-9)$. You can check any set of roots with graphing technology: graph the relation and read where it crosses the $x$-axis.

1. Rearrange so that one side is 0. Move every term to one side and write it in standard form, $a{x}^2+bx+c=0$.
2. Take out a common factor if there is one. If $a$ is negative, taking out $-1$ (or a negative common factor) makes the bracket easier to factor.
3. Factor the expression fully, using the methods from the Quadratic Expressions unit.
4. Set each factor that contains x equal to 0. A number factor such as the 3 in $3(x-6)(x+2)$ can never equal 0, so it gives no root.
5. Solve each small linear equation.
6. Verify each root in the original equation, and in a word problem, check that each root makes sense.

The perfect square row deserves a second look. Both factors of $(x-3)(x-3)=0$ give the same root, so the equation has only one distinct root, $x=3$. It is sometimes called a double root or two equal roots. On the graph of $y={(x-3)}^2$, the vertex sits on the $x$-axis at $(3,0)$: the parabola touches the axis there without crossing it. In the third lesson of this unit you will predict this case without factoring.

The patio equation $x^2+10x+16=40$ is not ready for the zero product property, because its right side is 40, not 0. Subtract 40 from both sides: $x^2+10x-24=0$. You need two numbers that multiply to $-24$ and add to $10$: they are $12$ and $-2$. So $(x+12)(x-2)=0$, which gives $x=-12$ or $x=2$. Both are correct algebra, but only one fits the situation. With $x=-12$ the sides would be $-10$ m and $-4$ m, and a length cannot be negative, so you reject that root. With $x=2$ the patio is 4 m by 10 m, and $4\times 10=40$ m², as required. In a word problem, state both roots, then say which one answers the question and why.