Factoring Quadratics

Today in class I presented a technique for factoring quadratics. Here are the steps. Remember that we’re starting from the right. This only works when the coefficient on the \(x^2\) term is 1.

  1. List the pairs of numbers that multiply to the constant (ignoring the sign!).
  2. The sign on the constant tells us whether we’re looking for a sum or a difference.
  3. The coefficient on the \(x\) term (ignoring the sign!) tells us what sum or difference we’re looking for.
  4. Now, using the sign on the \(x\) term, decide the signs for the factors.
    1. The higher factor will use the same sign.
    2. The lower factor will use the same sign if it’s a sum, and the opposite sign if it’s a difference.

Here are some examples.

\[x^2 + 7x + 12\]

  1. \(12 = 1 \times 12 = 2 \times 6 = 3 \times 4\)
  2. We’re looking for a sum. Our options are:
    1. \(1 + 12 = 13\)
    2. \(2 + 6 = 8\)
    3. \(3 + 4 = 7\)
  3. Our sum is 7. That means we want to use 3 and 4.
  4. We’re using +. So our factors are:
    1. \(x + 4\)
    2. \(x + 3\)

That means that \(x^2 + 7x + 12 = (x + 3)(x + 4)\).

\[x^2 – 2x – 63\]

  1. \(63 = 1 \times 63 = 3 \times 21 = 7 \times 9\)
  2. We’re looking for a difference. Our options are:
    1. \(63 – 1 = 62\)
    2. \(21 – 3 = 18\)
    3. \(9 – 7 = 2\)
  3. Our difference is 2. That means we want to use 7 and 9.
  4. We’re using -. So our factors are:
    1. \(x – 9\)
    2. \(x + 7\)

That means that \(x^2 – 2x – 63 = (x – 9)(x + 7)\).

Remember that the solutions will have the opposite signs to the factors, because solutions are values that make the factors equal to zero. So, in our first example, our solutions are \(x = {-3, -4}\), while in the second example, they’re \(x = {9, -7}\).

If this method doesn’t work, it means that at least one of the solutions of the quadratic expression isn’t an integer.

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