Outer Test#

Note that if we FOIL the expression \((x+m)(x+n)\), then we get:

\[(x+m)(x+n) = x^2 + nx + mx + mn = x^2 + \underbrace{(m+n)}_{b} x + \underbrace{(mn)}_{c}.\]

When factoring, we want to reverse this pattern. Given an \(x^2 + bx + c\), we’d like to find the numbers \(m\) and \(n\) so that \(x^2 + bx + c = (x+m)(x+n)\). To go from the right side to the left in the formula above, we need three things to work out:

  • The leading coefficient \(a\) needs to be 1. The trick only works on \(x^2 + bx + c\), not on \(7x^2 + bx + c\).

  • The sum of \(m\) and \(n\) needs to be \(b\)

  • The product of \(m\) and \(n\) needs to be \(c\).

In other words, we’re looking for factors of \(c\) that add up to \(b\).

Here’s an example.

Example 15

Let’s try to factor \(x^2 + 5x + 6\). This is a quadratic function in general form with \(a = 1\), \(b = 5\), and \(c = 6\). Since \(a = 1\), we can try out the factoring trick above. We just need to find factors of \(c = 6\) that add up to \(b = 5\). The factors of \(6\) that add up to \(5\) are \(2, 3\) (when you multiply them, you get 6, and when you add them, you get 5).

Therefore, \(x^2 + 5x + 6 = (x + 2)(x + 3)\).

Example 16

Click the Randomize button below to get a new example.

flowchart TD A[Start] --> B{Is it?} B -- Yes --> C[OK] C --> D[Rethink] D --> B B -- No ----> E[End]

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