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.

Start practice

Test yourself on solution by trinomial!

einstein

\( x^2+10x=-25 \)

Practice more now

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


Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
Test your knowledge

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


Do you know what the answer is?

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


Check your understanding

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)}


Test your knowledge

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)


Examples and exercises with solutions of square trinomial

Exercise #1

60βˆ’16y+y2=βˆ’4 60-16y+y^2=-4

Video Solution

Step-by-Step Solution

Let's solve the given equation:

60βˆ’16y+y2=βˆ’4 60-16y+y^2=-4 First, let's arrange the equation by moving terms:

60βˆ’16y+y2=βˆ’460βˆ’16y+y2+4=0y2βˆ’16y+64=0 60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0 Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

(aβˆ’b)2=a2βˆ’2ab+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2 This is done using the fact that:

64=82 64=8^2 So let's present the outer term on the right as a square:

y2βˆ’16y+64=0↓y2βˆ’16y+82=0 y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0 Now let's examine again the short factoring formula we mentioned earlier:

(aβˆ’b)2=a2βˆ’2abβ€Ύ+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 And the expression on the left side of the equation we got in the last step:

y2βˆ’16yβ€Ύ+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0 Let's note that the terms y2,82 \textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2 indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, meaning the middle term in the expression (underlined):

(aβˆ’b)2=a2βˆ’2abβ€Ύ+b2 (\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2 In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

y2βˆ’16yβ€Ύ+82=0↕?y2βˆ’2β‹…yβ‹…8β€Ύ+82=0 \textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0 And indeed it holds that:

2β‹…yβ‹…8=16y 2\cdot y\cdot8=16y So we can present the expression on the left side of the given equation as a difference of two squares:

y2βˆ’2β‹…yβ‹…8+82=0↓(yβˆ’8)2=0 \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

(yβˆ’8)2=0/yβˆ’8=Β±0yβˆ’8=0y=8 (y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y-8=\pm0\\ y-8=0\\ \boxed{y=8}

Let's summarize then the solution of the equation:

60βˆ’16y+y2=βˆ’4y2βˆ’16y+64=0↓y2βˆ’2β‹…yβ‹…8+82=0↓(yβˆ’8)2=0↓yβˆ’8=0↓y=8 60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}

So the correct answer is answer a.

Answer

y=8 y=8

Exercise #2

x2+10x=βˆ’25 x^2+10x=-25

Video Solution

Answer

x=βˆ’5 x=-5

Exercise #3

4x2=12xβˆ’9 4x^2=12x-9

Video Solution

Answer

x=32 x=\frac{3}{2}

Exercise #4

What is the value of x?

x4βˆ’x3=2x2 x^4-x^3=2x^2

Video Solution

Answer

x=βˆ’1,2,0 x=-1,2,0

Exercise #5

x2+144=24x x^2+144=24x

Video Solution

Answer

x=12 x=12

Check your understanding
Start practice