Question #f4266

Mar 11, 2015

Note: There are other ways of doing this (of thinking about it). This is one way.

Remember that $\left(a x + b\right) \left(c x + d\right) = a c {x}^{2} + \left(a d + b c\right) x + b d$

So the first two coefficients of the factors have to multiply to give us the number in front ${x}^{2}$.
Example: for factoring $6 {x}^{2} +$some other stuff,
the possibilities are $\left(6 x + b\right) \left(x + d\right)$ or $\left(2 x + b\right) \left(3 x + d\right)$.
(We don't need to worry about $\left(3 x + b\right) \left(2 x + d\right)$. because it's the same as the second choice with $b$ and $d$ exchanged.)

The constants have to multiply to give us the constant in the original expression.
Example: For $6 {x}^{2} + 11 x - 10$, the choices for how to end are $\left(a x + 10\right) \left(c x - 1\right)$ or the other way around $\left(a x - 10\right) \left(c x + 1\right)$
(There's a difference if $a \ne b$ because the 'minus' sign could go with either the $10$ or the $1$.)

OR $\left(a x + 5\right) \left(c x - 2\right)$ or $\left(a x - 5\right) \left(b x + 2\right)$

(Again the 'minus' could go with the $5$ or with the $2$.)

Finally, we also need $a d + b c =$ the coefficient of $x$,
which in this example is $11$. That's where the "trial" happens. We need to check each of the possibilities.

Here they are:
$\left(6 x + 10\right) \left(x - 1\right)$ or $\left(6 x - 10\right) \left(x + 1\right)$
$\left(2 x + 10\right) \left(3 x - 1\right)$ or $\left(2 x - 10\right) \left(3 x + 1\right)$

$\left(6 x + 5\right) \left(x - 2\right)$ or $\left(6 x - 5\right) \left(x + 2\right)$
$\left(2 x + 5\right) \left(3 x - 2\right)$ or $\left(2 x - 5\right) \left(3 x + 2\right)$

The faster you can do ad+bc (O+I, if you think FOIL),
the faster you can do the checking.

First line, I get $6$ and $10$, I'm not going to get $11$ from those.
Second line $2$ and $30$, no! I won't get $11$ from them.
Third line: $12$ and $5$, no good.
Fourth line $4$ and $15$ -- possible!. The difference is 11, so check further: $\left(2 x + 5\right) \left(3 x - 2\right)$ gives us $- 4$ and $+ 15$, and those add up to $11$. That's the one we want.

To be certain, multiply again:

$\left(2 x + 5\right) \left(3 x - 2\right) = 6 {x}^{2} - 4 x + 15 x - 10 = 6 {x}^{2} + 11 x - 10$.

So that works.

Remember you can't always factor by trial and error. ${x}^{2} + 2 x + 7$ won't factor using whole numbers.
The only possibilities would be $\left(x + 1\right) \left(x + 7\right)$ of $\left(x - 1\right) \left(x - 7\right)$. But neither of those work.
So ${x}^{2} + 2 x + 7$ is "prime" or irreducible" or "not factorable" -- depending on what you've been taught to call it.