Factoring trinomials

algebra factoring quadratic equation

The trick

To factor a trinomial like \(5x^2 -34x +24\), I was offered several times the following recipe.

  1. Move the leading coefficient to the constant term: \[ x^2 -34x +120 \]
  2. Factor the resulting trinomial: \[ x^2 -34x +120 = (x -30)(x -4) \]
  3. Add the moved coefficient back to the (two) variables \(x\): \[ (5x -30)(5x -4) \]
  4. One parenthesis has the very coefficient as common factor. Drop it: \[ 5x^2 -34x +24 = (x -6)(5x -4) \]

The recipe works, but the instructions are unfortunate. An explanation.
The underlying mathematical tool is a proper substitution. Remember: If a variable in an expression comes in a package wherever it shows up, as in \[ \tfrac{1 -2\sin x}{1 +2\sin x} +2\sin x\,, \] one can substitute a new variable \(z\) for the package (\(2\sin x\)) to get a less complicated, auxiliary expression \(\frac{1-z}{1+z}+z\). In our case \[ 5x^2 -34x +24 = 5x \cdot x -34x +24\,, \] we would like to substitute the package \((5x)\). To pack all occurrences of \(x\), we multiply the expression by \(5\), which changes its values! \[ 5\cdot (5x^2 -34x +24) = (5x)\cdot (5x) -34(5x) +120\,. \] Factor the auxiliary expression and replace \(z\) by its original meaning, \[ \begin{aligned} z^2 -34z +120 = (z -30)(z -4) &= 5(x -6)(5x -4)\,. \end{aligned} \] In a last step, we divide the expression by \(5\) to restore its values: \[ 5x^2 -34x +24 = (x -6)(5x -4)\,. \]

Quadratic equations

The tool can be used to complete the square in a quadratic equation. To solve \[ 6x^2 +x -3 = 0\,, \] we multiply by \(6\) and make the substitution \(z = 6x\). \[ \begin{aligned} z^2 +z -18 &= 0\\ z^2 +z &= 18 \end{aligned} \] Using \((z +\frac{1}{2})^2 = z^2 +z +\frac{1}{4}\), we get \[ \begin{aligned} 18 +\tfrac{1}{4} &= z^2 +z +\tfrac{1}{4}= (z +\tfrac{1}{2})^2\\ \pm \sqrt{\tfrac{73}{4}} &= z +\tfrac{1}{2}\\ z &= -\tfrac{1}{2} \pm \tfrac{1}{2}\sqrt{73}\,. \end{aligned} \] Therefore, since \(z = 6x\), the solutions are \(x = -\tfrac{1}{12} \pm \tfrac{1}{12}\sqrt{73}\).
Going back to the packing stage, one realizes that multiplying by \(24\) instead of \(6\) would benefit the whole calculation (why, and why not \(12\)?).