The trick
To factor a trinomial like 5x2−34x+24, I was offered several times the following recipe.
- Move the leading coefficient to the constant term: x2−34x+120
- Factor the resulting trinomial: x2−34x+120=(x−30)(x−4)
- Add the moved coefficient back to the (two) variables x: (5x−30)(5x−4)
- One parenthesis has the very coefficient as common factor. Drop it: 5x2−34x+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 1−2sinx1+2sinx+2sinx,
one can substitute a new variable z for the package (2sinx) to get a less complicated, auxiliary expression 1−z1+z+z. In our case 5x2−34x+24=5x⋅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⋅(5x2−34x+24)=(5x)⋅(5x)−34(5x)+120.
Factor the auxiliary expression and replace z by its original meaning, z2−34z+120=(z−30)(z−4)=5(x−6)(5x−4).
In a last step, we divide the expression by 5 to restore its values: 5x2−34x+24=(x−6)(5x−4).
Quadratic equations
The tool can be used to complete the square in a quadratic equation. To solve 6x2+x−3=0,
we multiply by 6 and make the substitution z=6x. z2+z−18=0z2+z=18
Using (z+12)2=z2+z+14, we get 18+14=z2+z+14=(z+12)2±√734=z+12z=−12±12√73.
Therefore, since z=6x, the solutions are x=−112±112√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?).