Squared Trinomial

πŸ†Practice solution by trinomial

More than once we have heard the teacher ask in class: "Who knows how to solve a quadratic equation without the formula?" We looked around to see who knew the answer, "Do you know what a trinomial is?" The teacher continued asking. We doubted and thought about what word this term could derive from and what a trinomial is. What does it really do? How does understanding about the trinomial benefit our mathematical knowledge? Does it expand the possibility of having greater mathematical efficiency? Or, in fact, might it be superfluous to include it in the ninth-grade curriculum?

In this article, we will try to answer these questions and even have fun with the properties of the trinomial that will help us quickly solve quadratic equations, to simplify fractions, to multiply and divide, to deal with fractions, even with the common denominator in fractions with variables in the numerator and in the denominator.

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Test yourself on solution by trinomial!

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\( 4x^2=12x-9 \)

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What is the origin of the word 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 22), 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:

X2+6X+8X^2+6X+8

Is a trinomial. Likewise, the form:

βˆ’X2βˆ’4X+8-X^2-4X+8

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

And, in a general form:

aX2+bX+caX^2+bX+c

It is a squared trinomial when the letters a,b,ca,b,c : are correct for any number we place in their place, except for 00

  • Note that, placing a 00 in place of the a will nullify the structure of the quadratic form
  • on the other hand, placing a 00 in place of the parameters bb or cc would turn the trinomial expression (three monomials) into one with only 22 monomials.

Basic concepts related to the quadratic equation and the trinomial

Quadratic form

An algebraic expression with an unknown XX or YY to the power of 22 is the highest exponent


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Coefficient

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


Parameter

A parameter is a letter other than XX or YY that is found within the expression instead of a number


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Monomial

Expression composed of the product or the quotient of a certain number with a variable such as: X5X5


Polynomial

Expression composed of the addition or subtraction of 22 or more monomials in which there are numbers and variables such as:

X+3X+3

4Xβˆ’24X-2 and others like this


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Where we will see squared trinomials in our math studies

  • 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.


Factoring Trinomials

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

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

We will obtain:

X2+3X+2X+2Γ—3= X^2+3X+2X+2 \times 3=

X2+(2+3)X+2Γ—3= X^2+(2+3)X+2 \times 3=

X2+5X+6 X^2+5X+6

To perform the opposite operation from trinomial to the original given multiplication, we will look for 22 numbers whose product gives said number without the X X.

In our case, it is 66 and the sum of both numbers results in the coefficient XX, which is the middle monomial of the trinomial -> in this case 55.

2Γ—3=6 2\times3=6

5=2+3 5=2+3

Then we factorize

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

and we will look for two numbers that comply with

aX2+bX+c aX^2+bX+c

So, in the following trinomial:

X2+8X+12 X^2+8X+12

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

8=2+6 8=2+6

12=6Γ—2 12=6\times2

Therefore, the factorization is:

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


Also in the following form:

X2βˆ’12X+20 X^2-12X+20

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

Consequently, we will obtain:

20=(10βˆ’)Γ—(2βˆ’) 20=(10-)\times(2-)

12βˆ’=(10βˆ’)+(2βˆ’) 12-=(10-)+(2-)

Therefore, the factorization is:

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


Do you think you will be able to solve it?

What happens when the product of the factors is negative?

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:

X2+5Xβˆ’6 X^2+5X-6

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

(Xβˆ’1)(X+6) (X-1)(X+6)

X2βˆ’7Xβˆ’8 X^2-7X-8

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

Because: 8βˆ’=(8βˆ’)(1)Β  8-=(8-)(1)Β 

and also: 7βˆ’=(8βˆ’)+1Β  7-=(8-)+1Β 

Therefore, the factorization is:

(X+1)(Xβˆ’8)Β  (X+1)(X-8)Β 


How to Solve Quadratic Equations Without a Formula

X2+11X+24=0Β  X^2+11X+24=0Β 

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

(X+8)(X+3)=0Β  (X+8)(X+3)=0Β 

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

This will happen when 8=XΒ Β  8 =XΒ Β 

When 0=3+XΒ  0= 3+XΒ 

So x=βˆ’3Β  x=-3Β 

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

x2+2x+1x2βˆ’xβˆ’2=(x+1)(x+1)(x+1)(xβˆ’2) \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:Β 

x+1xβˆ’2 \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.

1xβˆ’1Γ—x2βˆ’4x+3xβˆ’3= \frac{1}{x-1}\times\frac{x^2-4x+3}{x-3}=

1xβˆ’1Γ—(xβˆ’1)(xβˆ’3)xβˆ’3=1 \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.

x2+2x+1x2βˆ’6+9:x2+3x+2x2βˆ’xβˆ’6 \frac{x^2+2x+1}{x^2-6+9}:\frac{x^2+3x+2}{x^2-x-6}

(x+1)(x+1)(xβˆ’3)(xβˆ’3)Γ—(xβˆ’3)(x+2)(x+2)(x+1)=x+1xβˆ’3 \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:

1x2βˆ’2xβˆ’3βˆ’1xβˆ’3 \frac{1}{x^2-2x-3}-\frac{1}{x-3}
1(xβˆ’3)(x+1)βˆ’1xβˆ’3 \frac{1}{\left(x-3\right)\left(x+1\right)}-\frac{1}{x-3}
1βˆ’(x+1)(xβˆ’3)(x+1)=βˆ’x(xβˆ’3)(x+1) \frac{1-\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}=\frac{-x}{\left(x-3\right)\left(x+1\right)}


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How do you factor a trinomial in which the coefficient 2X is different from 1?

In general terms: aX2+bX+cΒ  aX^2+bX+cΒ 

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

Term A+ A + Term B=b B = b

Term AΓ— AΓ— Term B=ac B = ac
We will break down b 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:

2X2+5X+3= 2X^2+5X+3=

3+2=5,Β 3Γ—2=6 3+2=5 ,~ 3Γ—2=6

Therefore, we will obtain:

2X2+2X+3X+3= 2X^2+2X+3X+3=

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

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


Another system for factoring using the quadratic formula:

Thus, for example:

2X2+5X+3= 2X^2+5X+3=

We will obtain:

2(X2+2.5X+1.5)= 2(X^2+2.5X+1.5)=

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

X2+2.5X+1.5=0 X^2+2.5X+1.5=0

a=1,Β b=2.5,Β Β c=1.5 a=1,~b=2.5,~Β c=1.5

Now let's place it in the quadratic formula

βˆ’bΒ±b2βˆ’4ac2a= \frac{-b\pm\sqrt{b^2}-4ac}{2a}=

and we will arrive at:

X1=βˆ’1 X_1=-1, X2=βˆ’1.5 X_2=-1.5

a. We will start by factorizing X1 X_1 and X2 X_2:

a(Xβˆ’X1)(Xβˆ’X2) a(X-X_1)(X-X_2)

Therefore, the following factorization will be obtained:

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

  1. We multiply the 22 from the first parentheses to get whole numbers and it will be factorized:

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


Do you know what the answer is?

Substitution of numbers for letters in the trinomial

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

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

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

For example:

X2+2axβˆ’3a2 X^2+2ax-3a^2

We will discover that the product of the expressions is:

βˆ’aΓ—3a -aΓ—3a

is

βˆ’3a2 -3a^2

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

(X+3a)(Xβˆ’a) (X+3a)(X-a)

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

Thus, for example:

2X2+(2βˆ’a)Xβˆ’a 2X^2+(2-a)X-a

and we will obtain:

2X2+2Xβˆ’aXβˆ’a 2X^2+2X-aX-a

2X(X+1)βˆ’a(X+1) 2X(X+1)-a(X+1)

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

(X+1)(2Xβˆ’a) (X+1)(2X-a)


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