Vocabulary
- zero product property
This section has you taking your first steps into solving quadratic equations. You’re able to solve linear equations by isolating your variable through manipulating both sides of an equation. Quadratics, because they have both an $x^2$ and an $x$ term, require new strategies. The one we’ll focus on here is factoring.
Factoring quadratics involves breaking it apart into the two binomials that were multiplied to create it. For example, taking $x^2+3x+2$ and rewriting it as $(x+2)(x+1)$.
Like when we unwound standard form into vertex form, lets see how things in our binomials translate to standard form.
\[\begin{align*} (x + m)(x + n) &= x\cdot x + nx + mx +m\cdot n \\ &= x^2 + (m + n)x + mn \end{align*}\]Notice how $m$ and $n$ appear in both a sum and a product. That means if you can find a pair of numbers that add up to $b$ and multiply to $c$, then you can factor the quadratic. For the example above, $x^2+3x+2$, we needed two numbers that added up to 3 and whose product was 2. That means our numbers are 2 and 1, so our factored form is $(x+2)(x+1)$.
Negatives can be in the mix, too. For $x^2 - x - 12$ our two numbers have to add up to $-1$ and mutliply to $-12$. That would be $-4$ and $3$, giving us a factored form of $(x-4)(x+3)$.
When $a\neq 1$, we’ll have to expand on our trick from above. First, we’ll multiply $a$ and $c$. Then, we look for factors of that product that sum to $b$. We then split $x$-term into those numbers. Let’s look at $2x^2+11x+12$.
Now, we’re going to look at this new quadratic as two binomials, factoring what we can from each. Our first two terms have a $2x$ in common, and the second set of two has a 3.
\[2x(x+4) + 3(x + 4)\]Before we do this next step, I need to point out that there is no difference mechanically between what we have above and
\[2xy + 3y\]We have two terms with a common factor, so we can factor that out.
\[y(2x+3)\]And the same works with our somewhat factored quadratic.
\[(x+4)(2x+3)\]Let’s try another one.
Make sure you watch your negatives, and be sure that your binomials at the end match, like they do in step 4 above. If they don’t, go back and check for mistakes.
Great, you have a new version of a quadratic that seems just as complicated, if not more. And yes, if your goal was to get $x$ by itself, you’ll be sorely disappointed. But what this form is great when it’s equal to 0.
\[(x+2)(x+1)=0\\ x=-2 \quad\text{and}\quad x=-1\]The zero product property says that if two elements multiply to 0, then at least one of them has to be zero. So, we just need to find where each binomial is 0 to get out solutions.
Where this comes in handy is finding the $x$-intercepts, or zeros, of a quadratic function.
\[\begin{align*} y &= x^2 + 2x - 8 \\ y &= (x+4)(x-2) \end{align*}\]Our $x$-intercepts are at $(-4,0)$ and $(2,0)$, and one of their general uses is for finding positive and negative intervals. Since $x$-intercepts cross the $x$-axis, usually that comes with a sign switch.