# Factoring Trinomials: Part 1

You can use the distributive law to see that

$3\left(4n+5\right)=12n+15$,

and you can use FOIL to see that

$\begin{array}{l}\left(n+2\right)\left(n+3\right)=n\cdot n+n\cdot 3+2\cdot n+2\cdot 3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={n}^{2}+3n+2n+6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={n}^{2}+5n+6\end{array}$

You need to find two numbers which, when multiplied together yield the last term (the constant) of the trinomial – AND – which, when added together, yield the coefficient of the middle term (the ‘$b$’ in ${x}^{2}+bx+c$) of the trinomial.

## Factoring ${x}^{2}+bx+c$ when $b$ and $c$ are both positive

Example 1:

Factor ${n}^{2}+8n+15$.

The factoring can be done by finding two numbers whose sum is $8$ and product is $15$:

List pairs of numbers that have a product of $15$ and look for a pair that adds to $8$.

$\begin{array}{ccc}\left(1,15\right)& 1×15=15& 1+15=16\\ \left(3,5\right)& 3×5=15& 3+5=8\end{array}$

So, using $3$ and $5$ we can re-write the expression this way:

${n}^{2}+8n+15$

$={n}^{2}+3n+5n+15$ . . . . by re-writing $8n$ as $3n+5n$

$=\left({n}^{2}+3n\right)+\left(5n+15\right)$ …split the expression into two parts

$=n\left(n+3\right)+5\left(n+3\right)$ . . . .factor each part using the Distributive Property

$=\left(n+5\right)\left(n+3\right)$ . . . . . . . use the Distributive Property again to extract the factor $\left(n+3\right)$

So, the factored form of ${n}^{2}+8n+15$ is $=\left(n+5\right)\left(n+3\right)$.

Example 2:

Factor ${x}^{2}+37x+100$.

We need two numbers whose product is $100$ and sum is $37$.

$\begin{array}{l}100=\left(100\right)\left(1\right);100+1=101\\ 100=\left(50\right)\left(2\right);50+2=52\\ 100=\left(25\right)\left(4\right);25+4=29\\ 100=\left(20\right)\left(5\right);20+5=25\\ 100=\left(10\right)\left(10\right);10+10=20\end{array}$

It seems that $37$ never came up as a sum, so ${x}^{2}+37x+100$ cannot be factored (that is, it is an irreducible polynomial).

But do you see how you would factor ${x}^{2}+29x+100$?

See also Factoring: Parts 2, 3, and 4; factoring by grouping; and irreducible polynomials.