The summation of three monomials is known as a trinomial. Here, "tri" means "three". Trinomials contain three terms.

For example, x + 7y + 6z is a trinomial with three terms x, 7y and 6z.
x4y + 5z + 6 is a trinomial with three terms x4y, 5z and 6.
Trinomial Example

A Trinomial can be defined as a polynomial with only three terms. When operating with a trinomial, the following concepts are to be understood first:

Like Terms: These are terms with same variables.
For example: (2x, 7x), (8xy, -4xy), (2a2, -5a2)

Unlike Terms: These are terms with different variables.
For example: (4x, 8y), (3xy, 4x2), (5s,-6t)

Zero Property:
It is the combination of positive and negative like terms resulting in zero.

1x = x
-1x = -x
x0 = 1
x1 = x

While factorizing trinomials of the form x2 + ax + b, where a and b are integers, we have to express the trinomial as a product of two binomials.

The trinomial x2 + ax + b can be expressed as the product of two binomials of the form (x - A)(x - B), such that, A + B = a and AB = b.

Consider x2 + 4x + 3. This is a trinomial of the form x2 + ax + b where, a = 4 and b = 3.

x2 + 4x + 3 = (x + 1)(x + 3)

Observe that 3 + 1 = 4, which is the coefficient of x and 3 x 1 = 3 which is the constant term.

Let us consider another trinomial x2 - x - 6. This trinomial can be factored as (x + 2)(x - 3). Here, the sum of 2 and -3 is -1, which is the coefficient of x. The product of 2 and -3 is -6, which is the constant term.

While factoring trinomials of the form x2 + ax + b into binomials of the form (x + A)(x + B),
  • The sum of the constant terms of a binomial must be equal to the coefficient of x in the trinomial i.e. A + B = a
  • The product of the constant terms in the binomial must be equal to the constant term in the trinomial i.e AB = b
  • The sign of the constant term in the binomial is the same as the sign of the coefficient of x in the trinomial if the constant term in the trinomial is positive. i.e if b > 0, then A and B have the same sign as a.
  • If the constant term in the trinomial is negative, then the constant term in the binomial have opposite signs. i.e. if b is negative, then A and B have the opposite signs.
We can find the factors that satisfy all these conditions in a methodic way. Write down all the factors of the constant term in the trinomial and write down the sum of the factors. For example, consider the trinomial x2 + 5x - 24.

The factors of the constant term -24 and the sum of the factors are tabulated as follows:

Factors Sum of Factors
2, -12 2 + (-12) = -10
-2, 12 -2 + 12 = 10
3, -8 3 + (-8) = -5
-3, 8 -3 + 8 = 5
4, -6 4 + (-6) = -2
-4, 6 -4 + 6 = 2

-3 and 8 are two constants such that -3 x 8 is -24 and -3 + 8 is 5. The trinomial x2 + 5x - 24 = (x - 3)(x + 8)

Similarly, the trinomial of the form ax2 + bx + c, where a, b and c are real numbers and a ≠ 0 can be factored as a product of binomials with two more points to be taken care of
  • If the trinomial does not have any common factors, then the binomials will not have any common factors as well.
  • List the factors of a and c and then list all the combinations of the factors and check for the products.
When the leading coefficient is one, the factors can be formed by splitting the middle coefficient as the factors of the last term and pairing each with the variable present.

Examples of Factoring Trinomials


Given below are some of the examples that explain how to factorize trinomials.

Example 1:

Factorize 3x3 + 24x2 + 45x

Solution:

First find the GCF of the terms in the given trinomial. The GCF of the terms 3x3, 24x2, 45x is 3x

Now, factor the GCF out of all the terms.

3x3 + 24x2 + 45x = 3x(x2) + 3x(8x) + 3x(15)

= 3x(x2 + 8x + 15)

Now, factor the trinomial x2 + 8x + 15

Factors Sum
15, 1 16
3, 5 8

The sum of the factors 3 and 5 is 8. Hence the trinomial, x2 + 8x + 15 = (x + 3)(x + 5)

3x3 + 24x2 + 45x = 3x(x2 + 8x + 15)

= 3x(x+3)(x+5)

Hence, the trinomial 3x3 + 24x2 + 45x is factored to 3x(x+3)(x+5)

Example 2:

Factorize the trinomial 4y2 - 27y + 18.

Solution:

4y2 - 27y + 18 is a trinomial that does not have any common factors. And so, the binomial factors also do not have any common factors.

The constant term of the trinomial is +18. It is positive and so the signs of the constant terms of the binomial factors have the same sign. The coefficient of y is -27. It is negative and so the constant terms of the binomial factors have a negative sign.

Factors of 4 Factors of 18
1, 4 -1, -18
2, 2 -2, -9
4, 1 -3, -6

Using different combinations of these factors, let us write the binomial factors

(y - 1)(4y - 18) - There is a common factor

(y-2)(4y - 9)

(y - 3)(4y - 6) - There is a common factor

(2y - 1)(2y - 18) - There is a common factor

(2y - 2)(2y - 9) - There is a common factor

(2y - 3)(2y - 6) - There is a common factor

(4y - 1)(y -18)

(4y - 2)(y - 9) - There is a common factor

(4y - 3)(y - 6)

Now, we are left with three combinations

(y - 2)(4y - 9)

(4y - 1)(y - 18)

(4y - 3)(y - 6)

Calculate the middle terms for each of them

The middle term of (y - 2)(4y - 9) is -8y - 9y = -17y
The middle term of (4y - 1)(y - 18) is -72y - y = -73y
The middle term of (4y - 3)(y - 6) is -3y - 24y = -27y

Therefore, 4y2 - 27y + 18 = (4y - 3)(y - 6)
Given below are the steps to be followed while adding trinomials.

When adding trinomials,
  • Drop the parenthesis
  • Combine like terms
  • Combine unlike terms
  • Add numerical coefficients of like terms
  • Let the unlike terms remain as such

Examples on Adding Trinomials


Given below are some examples on adding trinomials.

Example 1:

Simplify (-2x2 + 4x + 1) + (5x2 - 3x - 6)

Solution:

(-2x2 + 4x + 1) + (5x2 - 3x - 6) = -2x2 + 4x + 1+ 5x2 - 3x - 6
= -2x2 + 5x2 + 4x - 3x + 1 - 6
= (-2 + 5) x2+ (4 - 3) x + (1 - 6)
= 3x2 + 1x + (-5)
= 3x2 + x - 5

Example 2:

Simplify (7x2 + 11x + 10) + (3x2 - x - 5)

Solution:

(7x2 + 11x + 10) + (3x2 - x - 5) = 7x2 + 11x + 10 + 3x2 - x - 5
= 7x2 + 3x2 + 11x - x + 10 - 5
= (7 + 3) x2+ (11 - 1) x + (10 - 5)
= 10x2 + 10x + 5
Given below are the steps to be followed while subtracting trinomials.

When subtracting trinomials,
  • Drop the parenthesis
  • Multiply the negative sign with all the terms of the second trinomial, so that all the signs of the terms of the second trinomial are changed to the opposite one. Hence “+” becomes “-” and “-” changes to “+”.
  • Combine like terms
  • Combine unlike terms
  • Subtract the numerical coefficients of like terms
  • Let the unlike terms remain as such.

Examples on Subtracting Trinomials


Given below are some examples on subtracting trinomials.

Example 1:

Simplify (-2x2 + 4x - 6) - (5x2 - 2x - 7)

Solution:

(-2x2 + 4x - 6) - (5x2 - 2x - 7) = -2x2 + 4x - 6 - 5x2 + 2x + 7
= -2x2 - 5x2 + 4x + 2x - 6 + 7
= (-2 -5) x2 + (4 + 2) x - (6 - 7)
= -7x2 + 6x - (-1)
= -7x2 + 6x + 1

Example 2:

Simplify (7x2 + 11x - 5) - (2x2 - x - 10)

Solution:

(7x2 + 11x - 5) - (2x2 - x - 10) = 7x2 + 11x - 5 - 2x2 + x + 10
= 7x2 - 2x2 + 11x + x - 5 + 10
= (7 - 2) x2 + (11 + 1) x - (5 - 10)
= 5x2 + 12x - (-5)
= 5x2 + 12x + 5

Multiplication can be done in two ways
  1. Horizontal Method
  2. Vertical Method

Horizontal Method of Multiplying Trinomials


The following steps have to be followed:

  • Choose the first and second trinomial (any order can be followed)
  • Arrange the trinomials vertically.
  • Distribute each term of the first polynomial with the second trinomial. Group the like terms.


Examples on Horizontal Method of Multiplying Trinomials


Given below are some examples on the horizontal method of multiplying trinomials.

Example 1:

Multiply (5x2 - 3x + 5) and (x2 + 5x - 2)

Solution:

Lets choose (5x2 - 3x + 5) as first term and (x2 + 5x - 2) as second term.
(5x2 - 3x + 5) $\times$ (x2 + 5x - 2) = 5x2 $\times$ (x2 + 5x - 2) - 3x $\times$ (x2 + 5x - 2) +5 $\times$ (x2 + 5x - 2)
= (5x2) (x2) + (5x2) (5x) - (5x2) (2) - (3x) (x2) - (3x) (5x) + (3x) (2) + (5) (x2) + (5) (5x) - (5) (2)
= 5x4 + 25x3 -10x2 - 3x3 -15x2 + 6x + 5x2 + 25x - 10
= 5x4 + 25x3 - 3x3 -10x2 -15x2 + 5x2 + 6x + 25x – 10
= 5x4 + (25 - 3) x3 - (10 + 15 - 5) x2 + (6 + 25) x – 10
= 5x4 + (22) x3 - (20) x2 + (31) x - 10
= 5x4 + 22x3 - 20x2 + 31x - 10
Example 2:

Multiply (4x2 + 2x - 1) and (x2 + 3x + 2)

Solution:

Lets choose (4x2 + 2x - 1) as first term and (x2 + 3x + 2) as second term.
(4x2 + 2x - 1) $\times$ (x2 + 3x + 2) = 4x2 $\times$ (x2 + 3x + 2) + 2x $\times$ (x2 + 3x + 2) - 1 $\times$ (x2 + 3x + 2)
= (4x2) (x2) + (4x2) (3x) + (4x2) (2) + (2x) (x2) + (2x) (3x) + (2x) (2) - (1) (x2) - (1) (3x) - (1) (2)
= 4x4 + 12x3 + 8x2 + 2x3 + 6x2 + 4x - x2 - 3x - 2
= 4x4 + 12x3 + 2x3 + 8x2 + 6x2 - x2 + 4x - 3x - 2
= 4x4 + (12 + 2) x3 + (8 + 6 - 1) x2 + (4 - 3) x - 2
= 4x4 + (14) x3 + (13) x2 + (1) x - 2
= 4x4 + 14x3 + 13x2 + x - 2

Vertical Method of Multiplying Trinomials


The following steps have to be followed:

  • Choose the first and second trinomial (any order can be followed)
  • Arrange them one below the other in vertical form.
  • Work from the right side to the left, take one term from the lower trinomial, and multiply it, again from right to left, with all terms on the top trinomial.
  • For each term in the lower trinomial, a row is formed underneath, stepping the rows off to the left as the process is worked from term to term in the lower trinomial.
  • Add up all the terms.


Examples on Vertical Method of Multiplying Trinomials


Given below are some examples on the vertical method of multiplying trinomials.

Example 1:

Multiply (x2 + 3x + 1) and (2x2 - 5x - 3)

Solution:

Lets choose (x2 + 3x + 1) as first term and (2x2 - 5x - 3) as second term.

x2 + 3x + 1
2x2 - 5x - 3
- 3x2 - 9x - 3
- 5x3 - 15x2 - 5x
2x4 + 6x3 + 2x2
2x4 + 1x3- 16x2 - 14x - 3

(x2 + 3x + 1) $\times$ (2x2 - 5x - 3) = 2x4 + 1x3- 16x2 - 14x - 3
= 2x4 + x3 - 16x2 - 14x - 3
Example 2:

Multiply (x2 + 2x + 1) and (x2 + 5x + 2)

Solution:

Lets choose (x2 + 2x + 1) as first term and (x2 + 5x + 2) as second term.

x2 + 2x + 1
x2 + 5x + 2
2x2 + 4x + 2
5x2 +10x2 +5x
x4 + 2x3 + x2
x4 + 7x3+ 13x2 + 9x + 2

(x2 + 2x + 1) $\times$ (x2 + 5x + 2) = x4 + 7x3+ 13x2 + 9x + 2

Division of a trinomial with a trinomial, results in a polynomial of degree less than 3.
There are different ways to do it.
  • By long term division
  • By factoring both the numerator and the denominator and simplifying.

Long Term Division of Dividing Trinomials


Take the dividend’s first term and divide it by the divisor’s first term and write the value as the quotient. In the next step, take the quotient obtained and multiply the divisor by it and place the respective value below the dividend keeping the like terms under each other. Subtract this from the dividend, take the reminder and any remaining terms from the dividend as the new dividend and repeat all the process until the degree of the reminder is less than the divisor.

Examples on Dividing Trinomials by Long Term Division


Given below are some examples on dividing trinomials by long term division.

Example 1:

$\frac{(8x^{2} - 16x + 8)}{(x^{2} - 2x + 1)}$
= 8

8
x2 - 2x + 1 ) 8x2 - 16x + 8
8x2 -16x + 8
0 0 0
Example 2:

$\frac{(5x^{2} + 10x + 5)}{(x^{2} + 2x + 1)}$
= 5

5
x2 + 2x + 1 ) 5x2 + 10x + 1
5x2 +10x + 5
0 0 0

Dividing Trinomials by Factoring Numerator and Denominator


In this method of dividing trinomials, first we have to factor both the numerator and the denominator separately and then divide them.

Example on Dividing Trinomials by Factoring Numerator and Denominator


Given below are some examples on dividing trinomials by factoring the numerator and the denominator.

Example 1:

$\frac{(8x^{2}-16x + 8)}{(x^{2} - 2x + 1)}$


Factor each part
(8x2-16x + 8) = 8x2 - 8x - 8x +8
= 8x (x - 1) -8 (x -1)
= (8x - 8) (x - 1)
= 8 (x - 1) (x - 1)

(x2 -2x + 1) = x2 - x - x + 1
= x (x - 1) -1 (x - 1)
= (x - 1) (x - 1)

So $\frac{(8x^{2} -16x + 8)}{(x^{2} - 2x + 1)}$ = $\frac{(8(x - 1) (x - 1))}{((x-1) (x-1))}$ = 8

Example 2:

$\frac{(5x^{2} + 10x + 5)}{(x^{2} + 2x + 1)}$

Factor each part
(5x2+10x + 5) = 5x2 + 5x + 5x +5
= 5x (x + 1) +5 (x +1)
= (5x + 5) (x + 1)
= 5 (x + 1) (x + 1)

(x2 +2x + 1) = x2 + x + x + 1
= x (x + 1) +1 (x + 1)
= (x + 1) (x + 1)

So $\frac{(5x^{2} + 10x + 5)}{(x^{2} + 2x + 1)}$ = $\frac{(5(x + 1) (x + 1))}{((x+1) (x+1))}$ = 5
There are two special products of trinomials which are the square of the sum and the square of the difference.

  • (a+b)2 = a2 + 2ab + b2
  • (a -b)2 = a2 - 2ab + b2
If the first and the last terms of a trinomial are perfect squares, the middle term has to be checked to confirm that it is a perfect square.

x2 + 8x + 16 The first and the last terms x2 and 16 are perfect squares.
x2 + 8x + 42
a2 + 2ab + b2
The square terms are rewritten and compared with the special products.
2ab = 2(x)(4) = 8x The middle term of the trinomial is verified.
x2 + 8x + 16 = (x + 4)(x + 4)
= (x+4)2
The Formula is applied to write the factors


Apply the sum or difference formula accordingly as the middle term is positive or negative.

Examples of Perfect Square Trinomials


Given below are some examples on perfect square trinomials.

Example 1:

Express 25x2 - 60xy +36y2 as a square of a Binomial.

Solution:

25x2 - 60xy +36y2 = (5x)2 - 2(5x)(6y) + (6y)2 (Here, a = 5x, b = 6y and the middle term is negative)

= (5x-6y)2

Example 2:

The area of a square is 4x2 + 28x + 49 sq.units. Find the side of the square as an expression of x.

Area of the square = 4x2 + 28x + 49

= (2x)2 + 2(2x)(7) + 72

= (2x + 7)2

Therefore, the side of the square = (2x + 7)units.