Factoring Trinomials: Part $4$

You can use the distributive law to see that

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

and you can use FOIL to see that

$\left(n+2\right)\left(n+3\right)={n}^{2}+5n+6$ .

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

Factoring $a{x}^{2}+bx+c,a\ne 1$

Things get a little trickier in this case. We need to find two numbers whose product is equal to the product of the leading coefficient and the constant and whose sum is equal to the coefficient of the $x$ term.

Example:

Factor $14{x}^{2}-37x+5$ .

Multiply the leading coefficient by the constant

$\left(14\right)\left(5\right)=70$

Find the factor pairs that multiply to $70$ and add to $-37$ .

$-2$ and $-35$

Replace the middle term.

$14{x}^{2}-2x-35x+5$
Factor common factors in pairs and use the Distributive Property.

$\left(14{x}^{2}-2x\right)-\left(35x-5\right)$

$2x\left(7x-1\right)-5\left(7x-1\right)$

Again, use the Distributive Property.

$\left(7x-1\right)\left(2x-5\right)$

See also Factoring: Parts $1$ , $2$ , and $3$ ; factoring by grouping ; and irreducible polynomials .