Squared Trinomial

Trinomial is a Greek word composed of two components:

• Tri - is three in Greek
• Nom - is partition - with unknown to the square (that is, to the second power)

Therefore: The squared trinomial is an algebraic expression composed of three terms, one with the unknown to the square (to the power $2$), the second with the unknown without exponentiation, and the third, numbers without unknowns or letters that are different from those carrying the unknown.

For example, the form:

$X^2+6X+8$

Is a trinomial. Likewise, the form:

$-X^2-4X+8$

is also, even though some of the coefficients of the unknowns are negative.

And, in a general form:

$aX^2+bX+c$

It is a squared trinomial when the letters $a,b,c$ : are correct for any number we place in their place, except for $0$

• Note that, placing a $0$ in place of the a will nullify the structure of the quadratic form
• on the other hand, placing a $0$ in place of the parameters $b$ or $c$ would turn the trinomial expression (three monomials) into one with only $2$ monomials.

An algebraic expression with an unknown $X$ or $Y$ to the power of $2$ is the highest exponent

The coefficient is a number that is written to the left of the variable.

A parameter is a letter other than $X$ or $Y$ that is found within the expression instead of a number

Expression composed of the product or the quotient of a certain number with a variable such as: $X5$

Expression composed of the addition or subtraction of $2$ or more monomials in which there are numbers and variables such as:

$X+3$

$4X-2$ and others like this

• In solving a quadratic equation without the formula
• In simplifying algebraic fractions that require converting a trinomial into a multiplication of factors to simplify
• In 4 mathematical operations with algebraic fractions to carry out the simplification or finding the appropriate common denominator, each trinomial structure is broken down into two factors

To this end, we will learn to break down every trinomial - a quadratic expression of three monomials into the multiplication of two polynomial factors.

To multiply the following expressions between parentheses let's return to the extended distributive property

$(X+3)(X+2)=$

We will obtain:

$X^2+3X+2X+2 \times 3=$

$X^2+(2+3)X+2 \times 3=$

$X^2+5X+6$

To perform the opposite operation from trinomial to the original given multiplication, we will look for $2$ numbers whose product gives said number without the $X$.

In our case, it is $6$ and the sum of both numbers results in the coefficient $X$, which is the middle monomial of the trinomial -> in this case $5$.

$2\times3=6$

$5=2+3$

Then we factorize

$(X+2)(X+3)$

and we will look for two numbers that comply with

$aX^2+bX+c$

So, in the following trinomial:

$X^2+8X+12$

We will find the numbers whose product is $12$ and their sum $8$. These are: $6$ and $2$, since:

$8=2+6$

$12=6\times2$

Therefore, the factorization is:

$(X+2)(X+6)$

Also in the following form:

$X^2-12X+20$

We will find the numbers whose product is $20$ and their sum $12$, knowing that both must be negative since the product is positive and the sum is negative.

Consequently, we will obtain:

$20=(10-)\times(2-)$

$12-=(10-)+(2-)$

Therefore, the factorization is:

$(X+2)(X+6)$

This indicates that these numbers have different signs and, when the sum is positive, it also suggests that the number with the larger defined value is positive.

For example:

$X^2+5X-6$

The numbers $6$ and $(-1)$, their sum is $5$ and their product $(-6)$. Therefore, we will factorize:

$(X-1)(X+6)$

$X^2-7X-8$

Let's find $2$ numbers whose product is $(-8)$ and whose sum is $(-7)$, and these are: $(-8)$ and $1$

Because: $8-=(8-)(1)$

and also: $7-=(8-)+1$

Therefore, the factorization is:

$(X+1)(X-8)$

$X^2+11X+24=0$

We will look for $2$ numbers whose product is $24$ and whose sum is $11$, and these are: $8, 3$ and we will obtain:

$(X+8)(X+3)=0$

To get the result $0$ it could be that the value of the expression in parentheses is $0=(8+X)$

This will happen when $8 =X$

When $0= 3+X$

So $x=-3$

And, in this way, simplifying algebraic fractions, we will factorize every trinomial and we can consider various possibilities for simplification:

$\frac{x^2+2x+1}{x^2-x-2}=\frac{\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}$

and we will arrive at: 

$\frac{x+1}{x-2}$

It cannot be simplified further due to the addition and subtraction operations between them.

Likewise, we can simplify in multiplication operations of fractions by decomposing the trinomial into a multiplication of two factors and simplifying as much as possible.

$\frac{1}{x-1}\times\frac{x^2-4x+3}{x-3}=$

$\frac{1}{x-1}\times\frac{\left(x-1\right)\left(x-3\right)}{x-3}=1$

Also in division operations: After converting the exercise to another of multiplicative inverse we will simplify as much as possible after factoring the trinomial.

$\frac{x^2+2x+1}{x^2-6+9}:\frac{x^2+3x+2}{x^2-x-6}$

$\frac{\left(x+1\right)\left(x+1\right)}{\left(x-3\right)\left(x-3\right)}\times\frac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x+1\right)}=\frac{x+1}{x-3}$

Similarly, to find the common denominator:

$\frac{1}{x^2-2x-3}-\frac{1}{x-3}$
$\frac{1}{\left(x-3\right)\left(x+1\right)}-\frac{1}{x-3}$
$\frac{1-\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}=\frac{-x}{\left(x-3\right)\left(x+1\right)}$

In general terms: $aX^2+bX+c$

We will look for two terms whose sum is $b$ and two factors whose product is $ac$

Term $A +$ Term $B = b$

Term $A×$ Term $B = ac$
We will break down $b$ into the sum of these two terms, which will allow us to factor the trinomial by taking the common factor out of the parentheses

For example:

$2X^2+5X+3=$

$3+2=5 ,~ 3×2=6$

Therefore, we will obtain:

$2X^2+2X+3X+3=$

$2X(X+1)+3(X+1)=$

$(2X+3)(X+1)$

Thus, for example:

$2X^2+5X+3=$

We will obtain:

$2(X^2+2.5X+1.5)=$

1. With the quadratic formula we will find the reset numbers of the trinomial within the parentheses:

$X^2+2.5X+1.5=0$

$a=1,~b=2.5,~ c=1.5$

Now let's place it in the quadratic formula

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

and we will arrive at:

$X_1=-1$, $X_2=-1.5$

a. We will start by factorizing $X_1$ and $X_2$:

$a(X-X_1)(X-X_2)$

Therefore, the following factorization will be obtained:

$2(X+1.5)(X+1)$

1. We multiply the $2$ from the first parentheses to get whole numbers and it will be factorized:

$(2X+3)(X+1)$

When substituting numbers in a trinomial with letters, it is essential that we do it according to the basic rules.

If $1=a$, we can carry out a shortened factorization

We will look for 2 terms whose product is $c$ and their sum $b$.

For example:

$X^2+2ax-3a^2$

We will discover that the product of the expressions is:

$-a×3a$

is

$-3a^2$

and their sum is $2a$, therefore, we will obtain

$(X+3a)(X-a)$

Also, when a is a number that is not $1$, we will factorize the coefficient $X$ into two terms whose product equals $ac$.

Thus, for example:

$2X^2+(2-a)X-a$

and we will obtain:

$2X^2+2X-aX-a$

$2X(X+1)-a(X+1)$

We will take out the common factor $X+1$ and obtain:

$(X+1)(2X-a)$