Factoring Trinomials

We will look for two numbers whose product is $a\times c$ and whose sum is $b$
We will ask ourselves: which number multiplied by which other will give us $a\times c$ or $​​c$ (if $a$ equals $1$).
and what plus what would add up to $b$.

In fact, we need to find a pair of numbers that meet these two conditions at the same time.

We can plot it as follows:

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$a$  The coefficient of the first term
$b$ The coefficient of the second term
$c$ The constant term

In the first step, we will use only addition to find the first solution, and then, we will use only subtraction to find the second.
Again, the factorization will look as follows:
$(x+solution \space one)(x+solution\space two)$
or with subtractions, depending on the solutions.

$ax^2+bx+c$

The trinomial represents an expression in which $x$ is squared, preceded by a coefficient (which can be positive or negative), but it must not be $0$ (sometimes the coefficient is equal to $1$ and therefore we will not see the $a$), to this term may be added or subtracted some other $bx$ when $b$ represents the coefficient (under the same conditions as $a$) and the independent variable (number $c$) is added or subtracted.
Regardless of whether the coefficients of the terms are positive or negative, as long as they appear in the form of a trinomial, the exercise will be called "trinomial".

We will look for two numbers whose product is $a*c$ and whose sum is $b$
We will ask ourselves: which number multiplied by which other number will give us $a\times c$ or $​​c$ (if $a$ equals $​​1$).
and what plus what would add up to $b$.

In fact, we have to find a pair of numbers that meet these two conditions at the same time.

We can plot it as follows:

AC Method:
We will find all the numbers whose products are $a\times c$ and write them down.
Then, we will see which pair of numbers among those we found will result in $B$.
The two numbers that meet both conditions are the solutions to the trinomial.

Important

• If A were different from $1$, it would appear before the parentheses and then there would be a multiplication.
• If any of the solutions or both were negative, we would not add them to the $X$ but subtract them instead.

$x^2+8x+12$
Let's find all the numbers whose products are $12$ (and remember them in negative as well)
we will obtain:
$12,1$
$2,6$
$3,4$
Now let's see which pair of numbers among those we already found will give us a total of $8$
The pair that meets both conditions is $2,6$.
Let's write the factorization:
$(x+2)(x+6)$

$x^2+4x+4=$

Let's find our parameters:
$a$    The coefficient of the first term $1$
$b$   The coefficient of the second term $4$
$c$  The constant term $4$

First, we will place them in the formula with the plus sign and it will give us:
$\frac{-4+\sqrt{4^2-4\times 1\times 4}}{2\times 1}=$
$\frac{-4+\sqrt{16-16}}{2}=$
$\frac{-4+\sqrt{0}}{2}=$
$\frac{-4}{2}=-2$
We will place them in the formula with the minus sign and we will get:
$\frac{-4-\sqrt{0}}{2}=$
$-\frac{4}{2}=-2$

We get the same answer.
The factorization is:
$(x-2)(x-2)$

If you are interested in this article, you might also be interested in the following articles:

• Factorization
• The uses of factorization
• Factorization according to short multiplication formulas
• Factorization through the extraction of the common factor outside the parentheses
• Factorization of algebraic fractions
• Addition and subtraction of algebraic fractions
• Simplification of algebraic fractions
• Multiplication and division of algebraic fractions
• Solving equations through factorization

In the Tutorela blog, you will find a variety of articles about mathematics.