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

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

But how can you start with the answer and find the factors?

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.