More on factoring

More advanced quadratic factoring

Even if the regular method of quadratic factoring does not work, a quadratic may still be factorable if it has roots. Therefore we check the discriminant \(\Delta=b^2-4ac\) to see if it is positive. Then we calculate the roots with the quadratic formula, and we can write the factored form directly. Consider the quadratic \(3x^2-2x-7\), in which \(a=3, b=-2\) and \(c=-7\). The discriminant is \[\Delta=(-2)^2-(4)(3)(-7)=4+84=88\] Therefore, the roots are \[x=\dfrac{-(-2)\pm\sqrt{88}}{(2)(3)}=\dfrac{2\pm2\sqrt{22}}{(2)(3)}=\dfrac{1\pm\sqrt{22}}{3}\] Hence, the factored quadratic is \[3\left(x-\dfrac{1+\sqrt{22}}{3}\right)\left(x-\dfrac{1-\sqrt{22}}{3}\right)\]

Change of variable

In certain situations, a change-of-variable approach makes the pattern appear more clearly. Consider the expression \[2\sin^2 x-7\sin x+3\] Now let \(y=\sin x\) and replace that in the expression, which then becomes \[2y^2-7y+3\] We are now dealing with a regular quadratic and we can factor it as \[(2y-1)(y-3)\] After that, we substitute back with the original variable, and the factored expression becomes \[(2\sin x-1)(\sin x-3)\]

Quadratics in two (or more) variables

Consider the expression \[4x^2-3xy-y^2\] If we take \(y^2\) as a common factor (assuming it is not zero) we get \[y^2\left[4\left(\dfrac{x}{y}\right)^2-3\dfrac{x}{y}-1\right]\] If the pattern is not obvious, let \(z=\frac{x}{y}\), and the expression becomes \[y^2\left[4z^2-3z-1\right]\] which can then be factored as \[y^2(4z+1)(z-1)\] When we substitute back, we get \[y^2\left(4\dfrac{x}{y}+1\right)\left(\dfrac{x}{y}-1\right)\] Then we can distribute \(y^2\) back into the brackets (one \(y\) into each bracket) and we get \[(4x+y)(x-y)\] Note that this expression is also equivalent to the original even if \(y=0\). In other words, we can factor any quadratic expression in any number of variables as long as it is quadratic in all of those variables. Factoring is then based on the coefficients only.

Quadratics in higher powers

A simple change-of-variable technique can turn higher polynomials into simple quadratics. Consider \[4x^8-x^4-3\] Instead of attempting to factor this by assuming it has a rational root, we can let \(y=x^4\); then the expression becomes \[4y^2-y-3\] and we can factor it as \[(4y+3)(y-1)\] Substituting back, the factored expression is then \[(4x^4+3)(x^4-1)\] Then we can factor the second set of brackets as a difference of squares \[(4x^4+3)(x^2+1)(x^2-1)\] and then again \[(4x^4+3)(x^2+1)(x+1)(x-1)\] Attempting to factor the polynomial by searching for rational roots takes much longer and requires an additional step of factoring by grouping. Note that a change of variable is not absolutely necessary, but it helps.

Using identities

This can be tricky and is not always guaranteed to work, but it is worth considering when trigonometric, exponential, or logarithmic functions are involved. Consider the following (complicated) example \[1-3\cos 2x-\sin 2x\] Using the double-angle identities and the Pythagorean identity, we get \[\begin{aligned} &\phantom{={}}\sin^2 x+\cos^2 x-3(\cos^2 x-\sin^2 x)-2\sin x\cos x \\ &=\sin^2 x+\cos^2 x-3\cos^2 x+3\sin^2 x-2\sin x\cos x \\ &=4\sin^2 x-2\sin x\cos x-2\cos^2 x \\ &=2(2\sin^2 x-\sin x\cos x-\cos^2 x) \end{aligned}\] At this point, if the pattern is not obvious, we let \(y=\sin x\) and \(z=\cos x\). The expression then becomes \[2(2y^2-yz-z^2)\] which is factored as a quadratic in two variables into \[2(2y+z)(y-z)\] When we substitute back we get the final factored form \[2(2\sin x+\cos x)(\sin x-\cos x)\]