# How do you factor ?

Jul 24, 2015

It depends.

#### Explanation:

Here's an incomplete list of pointers:

(1) Do all the terms have a common factor?

If so, separate out the common factor first...

$2 {x}^{2} + 8 x + 8 = 2 \left({x}^{2} + 4 x + 4\right)$

$12 {x}^{2} y - 4 x {y}^{2} + 8 x y = 4 x y \left(3 x - y + 2\right)$

(2) Do all the coefficients add up to $0$?

e.g. $f \left(x\right) = {x}^{2} + 3 x - 4$

Here $f \left(1\right) = 1 + 3 - 4 = 0$. Hence $\left(x - 1\right)$ is a factor of $f \left(x\right)$

(3) Do the coefficients of the even degree terms minus the coefficients of the odd degree terms add up to $0$?

e.g. $f \left(x\right) = {x}^{2} - 3 x - 4$

Here $f \left(- 1\right) = 1 - \left(- 3\right) - 4 = 0$. Hence $\left(x + 1\right)$ is a factor of $f \left(x\right)$

(4) Is the polynomial a difference or sum of squares or cubes?

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

e.g. $4 {x}^{2} - 9 = \left(2 x - 3\right) \left(2 x + 3\right)$

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

e.g. $8 {x}^{6} - 27 = \left(2 {x}^{2} - 3\right) \left(4 {x}^{4} + 6 {x}^{2} + 9\right)$

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Note that ${a}^{2} + {b}^{2}$ is only reducible to linear factors with complex coefficients.

${a}^{2} + {b}^{2} = \left(a - i b\right) \left(a + i b\right)$

(5) Is the polynomial recognisable as a perfect square trinomial?

${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

e.g. $4 {x}^{2} - 12 x + 9 = {\left(2 x - 3\right)}^{2}$

(6) Is the polynomial easily factorisable as $\left(x + a\right) \left(x - b\right)$ or similar?

$\left(x + a\right) \left(x - b\right) = {x}^{2} + \left(a - b\right) x - a b$

e.g. ${x}^{2} + x - 12 = \left(x + 4\right) \left(x - 3\right)$ based on the observation that $4 \cdot 3 = 12$ and $4 - 3 = 1$

(7) Check the discriminant for a quadratic

For a quadratic in the form $f \left(x\right) = a {x}^{2} + b x + c$, check the value of the discriminant $\Delta = {b}^{2} - 4 a c$.

If $\Delta > 0$ and is a perfect square, then it may be simplest to use some version of the AC Method or similar to factor the quadratic as it does have rational factors.

If $\Delta > 0$ is not a perfect square, then use the quadratic formula to find the roots ${x}_{1}$ and ${x}_{2}$ of $f \left(x\right) = 0$ and hence factor $f \left(x\right)$ as $a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

If $\Delta = 0$, you have a perfect square trinomial. Look at (5) above.

If $\Delta < 0$, $f \left(x\right)$ has no linear factors with real coefficients. If you allow complex coefficients you can still use the quadratic formula, etc, but the result will be linear factors with complex coefficients.