As we start studying algebra, we come across polynomials. These are the algebraic expressions containing variables and coefficients which are combined together by using algebraic operators, such as - plus (+), minus (-), multiply (x) and divide (/). There are following 4 types of expressions in mathematics :
(1) Monomial - An expression which is made up of only one term is known as monomial. e.g. x$^{3}$, 7ab.
(2) Binomial - An expression containing at the most two terms is termed as a binomial. e.g. 2x - 5, m$^{2}$ + 4mn.
(3) Trinomial - Trinomial is a type of expression which is composed of three terms. e.g. a - b + 4, x$^{3}$ - 3xy - y$^{3}$.
(4) Polynomial - An expression containing many terms and is of the following form is known as a polynomial of degree n -
f (x) = $a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_{0}$
where, $a_{n} \neq 0$
Let us study about trinomials in more detail in this page. Trinomial is an expression containing 3 unlike terms. The degree of the trinomial is the highest degree in the expression. If the highest degree of all variables put together is 2, then it is called quadratic and if it is 3, then it is a cubic function. Factoring is a reverse process of the FOIL method. Factoring trinomial means to reduce polynomial into binomials.
Below you can see the steps:
Step 1: Factor out common factor, if possible.
Step 2: Factorized the trinomial using suitable method.
Step 3: Check whether the factors themselves can be factored.
Step 4: Verify the results by multiplying the factors.
Use the below widget to factor :
Quadratic trinomial can be factored using the following three methods:
- Factoring by grouping
- Completing the square
- Using quadratic formula.
Some of the identities used while factoring trinomials are
- a^{2 }+ 2ab + b^{2 }= (a + b)^{2}
- a^{2} - 2ab + b^{2} = (a - b)^{2}
- a^{2 }- b^{2 }= (a + b)(a - b)
- x^{2} + x(a + b) + ab=(x + a)(x + b)
Factoring Trinomials 'x Method'
A quadratic trinomial is a trinomial of three terms. The "x Method" of factoring helps us to see the possible combination of numbers that will equal to a given number when multiplied and added together.
Factoring Trinomials by Grouping
Let us find the factors of x^{2} + 9x + 8.
Given x^{2} + 9x + 8
Step 1:
Multiply the leading coefficient, a and constant term, c
Here, a = 1 and c = 8
=> ac = 1 * 8 = 8
Find the factors of ac.
1 * 8, 2 * 4
Find out the factor pair that adds to the middle term, b (coefficiant of x)
1 + 8 = 9
Step 2:
x^{2} + 9x + 8 = x^{2} + x + 8x + 8
Group the first two terms together and group the last two terms together.
=> x
^{2} + 9x + 8 = x(x + 1) + 8(x + 1)
= (x + 8)(x + 1)
Therefore, factors of x
^{2} + 9x + 8 are (x + 8) and (x + 1).
Factoring Trinomials Box Method
Let us factor 4x
^{2} + 11x - 3 by box method.
Given 4x
^{2} + 11x - 3
Step 1:
Create a box of 2 x 2
Step 2:
In the top left corner, put the first term and in the bottom right corner put the last term of polynomial.
Step 3:
Multiply first and last term of given polynomial
=> 4x^{2} * -3 = -12x^{2}
Find two factors of -12x^{2}, that when added together they will give the middle term, 11x.
=> Factors are 12x and -x.
Step 4:
Put factors into the open boxes.
Step 5:
Factor the terms in each row and in each column
Step 6:
The sum of the factors for the rows and the sum of the factors for the columns are the required factors.
=> (4x - 1) and (x + 3) are the factors of given polynomial.
Perfect square trinomial is the product of two identical binomials. During factoring, the first term and last term of the perfect square are perfect squares
and the middle term is 2 times the square root of first terms times and
square root of last terms.
Quadratic trinomial is a trinomial containing three terms with degree 2. The general form of quadratic trinomial is ax
^{2} + bx + c, a $\neq$ 0. Trinomial can be factorized by factor theorem or by using quadratic formula.
Factoring Trinomials Formula
Trinomial can be factorized by using quadratic formula:
x = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$Let us find the factors of 2x
^{2} - 5x - 3.
Given 2x
^{2} - 5x - 3
Factorized the quadratic
2x
^{2} - 5x - 3 = 2x
^{2} - 6x + x - 3
= 2x(x - 3) + (x - 3)
= (2x + 1)(x - 3)
Therefore, factors of 2x
^{2} - 5x - 3 are (2x + 1) and (x - 3).
A cubic trinomial is a polynomial of degree 3. The general form of cubic polynomial is, ax
^{3} + bx
^{2} + cx + d, where a, b, c and d are real coefficients and a $\neq$ 0. But in this case we considered cubic polynomials with three terms. Lets see with the help of example how to factoring cubic trinomial.
Let us factorize x
^{3} + x - 2
Given cubic trinomial is x
^{3} + x - 2.
Step 1:Find the first root by trial and error method
Factors of 2 = $\pm$ 1, $\pm$ 2
Put x = 1 in x
^{3} + x - 2
=> x
^{3} + x - 2 = 1
^{3} + 1 - 2 = 0
So, x = 1 is a root of x
^{3} + x - 2.
Step 2:For other roots, use synthetic division method.
Factors of x
^{2} + x + 2 are (x + $\frac{-1 \pm \sqrt{7}i}{2}$) (x - $\frac{-1 \pm \sqrt{7}i}{2}$).
=> Factors of x
^{3} + x - 2 are (x - 1)(x +
$\frac{-1 \pm \sqrt{7}i}{2}$ ) and (x -
$\frac{-1 \pm \sqrt{7}i}{2}$).
The general form of trinomial in two variables is x
^{2} + bxy + cy
^{2}. This trinomial has two variables, x and y.
From factorization theorem, factors of this trinomials are (x + my)(x + ny)
=> x^{2} + bxy + cy^{2} = x^{2} + (m + n)xy + (mn)y^{2} = (x + my)(x + ny)
Let us factor the trinomial x
^{2} - 6xy + 8y
^{2} Given x
^{2} - 6xy + 8y
^{2} ....................(i)
We know that, x
^{2} + bxy + cy
^{2} = x
^{2} + (m + n)xy + (mn)y
^{2} .............(ii)
From (i) and (ii)
=> x
^{2} + (m + n)xy + (mn)y
^{2} = x
^{2} - 6xy + 8y
^{2} Comparing coefficient of x
^{2}, we have
1 = 1
Comparing coefficients of xy, we have
=> m + n = -6 ................(iii)
Comparing coefficient of y
^{2}, we have
mn = 8 .......................(iv)
The two numbers -2 and -4 satisfy both conditions given in (iii) and (iv):
=> m = -4 and n = -2
=> x
^{2} - 6xy + 8y
^{2} = (x + (-4)y)(x + (-2)y) = (x - 4y)(x - 2y).
A trinomial equation is a polynomial equation involving 3 terms. Trinomial equations can be solved by solving their factors. Each factor becomes a separate equation and gives the value of trinomial. Usually, we deal with trinomials in the from of quadratic.
Let us solve for x, 3x^{2} + 5x - 2 = 0.
Given trinomial equation is 3x^{2} + 5x - 2 = 0
Last term in the trinomial = -2
Since cofficient of x^{2 } $\neq$ 0
Find the factors of ac = 3 * -2 = -6 (where a = coefficient of x^{2} and c = constant term)
-1 * 6, 2 * -3, 1 * -6, -2 * 3
Choose two factors of -6, so that their sum is the coefficient of x.
=> 6 - 1 = 5
Now,
3x^{2} + 5x - 2 = 3x^{2} + 6x - x - 2
= 3x(x + 2) - (x + 2)
= (3x - 1)(x + 2)
(3x - 1) and (x + 2) are factors of 3x^{2} + 5x - 2.
Solve for x, set each binomial equal to zero.
=> 3x - 1 = 0 or x + 2 = 0
=> x = $\frac{1}{3}$ or x = -2
Therefore, values of x are -2, $\frac{1}{3}$.
Given below are some examples:
Solved Examples
Question 1: Factorize x
^{2} - 9x - 70
Solution:
Given x^{2} - 9x - 70
x^{2} - 9x - 70 = x^{2} - 14x + 5x - 70
= x(x - 14) + 5(x - 14)
= (x + 5)(x - 14)
Factors of x^{2} - 9x - 70 are (x + 5) and (x - 14).
Question 2: Factorize x
^{2} - 18x + 81
Solution:
Given x^{2} - 18x + 81
x^{2} - 18x + 81 = x^{2} - 9x - 9x + 81
= x(x - 9) - (x -9)
= (x - 1)(x - 9)
Factors of x^{2} - 18x + 81 are (x - 1) and (x - 9).
Given below are some of the practice problems in factoring trinomials.
Practice Problems
Question 1: Factorize x^{2} + 19x + 90
Question 2: Factorize x^{2} + 6x - 55