In this chapter, we will learn to factorize quadratic trinomial with solved examples.
At the end of the chapter, problems are also given for practice.
Before moving on with the chapter, let us first understand the basics.
What are quadratic trinomials ?
The math expression in the form of \mathtt{ax^{2} +bx\ +c} are called quadratic expressions.
It is called trinomial because it contains total of three terms separated by addition/subtraction sign.
Given below are some examples of quadratic equation.
\mathtt{\Longrightarrow \ 5x^{2} +6x+11}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -2x+1}\\\ \\ \mathtt{\Longrightarrow \ -11x^{2} +10x-21}
Factorizing quadratic trinomials
We will understand the process with example.
Consider the below quadratic expression;
\mathtt{\Longrightarrow \ 2x+x^{2} -63}
To solve the expression follow the below steps;
(a) First arrange the trinomial from highest to lowest degree.
\mathtt{\Longrightarrow \ x^{2} +2x-63}
(b) Multiply the coefficient of \mathtt{x^{2}} with constant term.
So, multiplying the first & last term we get;
\mathtt{\Longrightarrow \ 1\ \times \ -63}\\\ \\ \mathtt{\Longrightarrow \ -63}
(c) Find the coefficient of x.
Here coefficient of x is 2.
(d) Find two numbers which on multiplication produce number -63 and on addition produce number 2.
Let a & b are the two numbers.
So;
a x b = -63
a + b = 2
We can find both the number by finding individual factors of number 63.
Factors of 63 = 1, 3, 7, 9, 21, 63
If we use factors 9 & -7, our both the condition will be satisfied.
9 x -7 = -63
9 +(-7) = 2
Hence, we got the value of two numbers a & b.
(e) Expand the middle part of quadratic expression using the above numbers a & b and then factorize by grouping.
\mathtt{\Longrightarrow \ x^{2} +2x-63}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +( 9-7) x-63}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +9x-7x-63}\\\ \\ \mathtt{\Longrightarrow \ x( x+9) -7( x+9)}\\\ \\ \mathtt{\Longrightarrow \ ( x+9)( x-7)}
Hence, the quadratic expression has been factorized to (x+9)(x-7).
I hope you understood the above process. You have to follow exact same steps to factorize the quadratic expression.
Given below are some solved examples for further clarity.
Example 01
\mathtt{\Longrightarrow \ 2x^{2} +7x-9}
Solution
Follow the below steps;
(a) Arrange the quadratic expression from highest to lowest degree.
The expression is already arranged like that.
\mathtt{\Longrightarrow \ 2x^{2} +7x-9}
(b) Multiply coefficient of \mathtt{x^{2}} and constant.
Multiply first and last term.
\mathtt{\Longrightarrow \ 2\ \times \ -9}\\\ \\ \mathtt{\Longrightarrow \ -18}
(c) Identify the coefficient of x
Coefficient of x is 7.
(d) Find two numbers a & b which on multiplication produce -18 and on addition produce 7.
This can be found by finding factors of 18.
Factors of 18 = 1, 2, 3, 6, 9, 18
The suitable values of a & b are 9 and -2.
Checking of the values are suitable.
9 x -2 = 18
9 + (-2) = 7
(e) Expand the middle part of trinomial using values of a & b and then factorize.
\mathtt{\Longrightarrow \ 2x^{2} +7x-9}\\\ \\ \mathtt{\Longrightarrow \ 2x^{2} +\ ( 9-2) x-9}\\\ \\ \mathtt{\Longrightarrow \ 2x^{2} +9x-2x-9}\\\ \\ \mathtt{\Longrightarrow \ 2x^{2} -2x\ +9x-9}\\\ \\ \mathtt{\Longrightarrow \ 2x( x-1) +9( x-1)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)( 2x+9)}
Hence, the above expression is the factorized form of quadratic equation.
Example 02
Factorize the quadratic equation
\mathtt{\Longrightarrow \ 45+14x+x^{2}}
Solution
Follow the below steps;
(a) Arrange the quadratic expression from high to low degree.
\mathtt{\Longrightarrow \ x^{2} +14x\ +\ 45}
(b) Multiply coefficient of \mathtt{x^{2}} and constant term.
\mathtt{\Longrightarrow \ 1\ .\ 45}\\\ \\ \mathtt{\Longrightarrow \ 45}
(c) Find coefficient of x.
Here coefficient of x is 14.
(d) Find two numbers a & b which on multiplication produce 45 and on addition produce 14.
This can be done by finding factors of 45.
Factor of 45 = 1, 3, 5, 9, 15, and 45
The two numbers are 9 and 5.
Checking the numbers;
9 x 5 = 45
9 + 5 = 14
(e) Expand the middle part of quadratic expression using values of a & b and then factorize.
\mathtt{\Longrightarrow \ x^{2} +14x\ +\ 45}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +( 9+5) x+45}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +9x\ +5x\ +\ 45}\\\ \\ \mathtt{\Longrightarrow \ x( x+9) +5( x+9)}\\\ \\ \mathtt{\Longrightarrow \ ( x+5) .( x+9)}
Example 03
Factorize the quadratic expression
\mathtt{\Longrightarrow \ -8x^{2} -15x\ +\ 2}
Solution
Follow the below steps;
(a) Arrange the expression from highest degree to lowest degree.
The quadratic trinomial is already arranged like that.
\mathtt{\Longrightarrow \ -8x^{2} -15x\ +\ 2}
(b) Multiply coefficient of \mathtt{x^{2}} and constant.
i.e. multiply first and last terms.
\mathtt{\Longrightarrow \ -8\ \times \ 2}\\\ \\ \mathtt{\Longrightarrow \ -16}
(c) Find the coefficient of x.
Here coefficient of x is -15
(d) Find the values of a & b such that on multiplication they produce -16 and addition -15.
To get these numbers, first find factors of 16.
Factors of 16 = 1, 2, 4, 8, and 16
The two numbers are -16 + 1.
Checking the numbers;
-16 x 1 = -16
-16 + 1 = -15
(e) Expand the middle values of trinomial and factorize using grouping method.
\mathtt{\Longrightarrow \ -8x^{2} -15x\ +\ 2}\\\ \\ \mathtt{\Longrightarrow \ -8x^{2} -16x+x\ +2}\\\ \\ \mathtt{\Longrightarrow \ \left( -8x^{2} -16x\right) +( x\ +2)}\\\ \\ \mathtt{\Longrightarrow \ -8x( x+2) +( x+2)}\\\ \\ \mathtt{\Longrightarrow \ ( x+2) \ ( -8x+1)}
Hence, the quadratic equation is factorized to (x+2) (-8x+1)
Example 04
Factorize the below quadratic trinomial.
\mathtt{\Longrightarrow \ x^{2} +6x-10=30}
Solution
First simplify the expression so that it contain only three elements.
\mathtt{\Longrightarrow \ x^{2} +6x-10=30}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +6x-10-30\ =0}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +6x-40=0}
Now follow the below steps;
(a) Arrange the expression from highest to lowest degree.
The expression is already arranged like that.
\mathtt{\Longrightarrow \ x^{2} +6x-40=0}
(b) Multiply the coefficient of \mathtt{x^{2}} with the constant.
i.e. Multiply the first and last terms.
\mathtt{\Longrightarrow \ 1\times \ -40}\\\ \\ \mathtt{\Longrightarrow \ -40}
(c) Find the coefficient of x.
Coefficient of x is 6.
(d) Find two numbers a & b such that their multiplication produce -40 and addition produce 6.
This can be found by finding factors of 60.
Factors of 60= 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60
From the above factors, find the two numbers that satisfy the condition.
The two numbers are 10 and -4.
i.e. a= 10 and b = -4.
Checking the numbers;
10 x -4 = -40
10 + (-4) = 6
Hence, these selections are correct.
(e) Expand the middle values of trinomial using a & b and then factorize.
\mathtt{\Longrightarrow \ x^{2} +6x-40=0}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +( 10-4) x-40\ =0}\\\ \\ \mathtt{\Longrightarrow \ x^{2} +10x-4x-40\ =\ 0}\\\ \\ \mathtt{\Longrightarrow \ x( x+10) -4( x+10) =0}\\\ \\ \mathtt{\Longrightarrow \ ( x-4)( x+10) \ =0}
Hence, the quadratic expression has been factorized to given above expression.
Example 05
Factorize the below expression.
\mathtt{\Longrightarrow \ 7x^{2} -15x+8}
Solution
Follow the below step;
(a) Arrange the quadratic expression from highest to lowest order.
The expression has already been arranged like that.
(b) Multiply coefficient of \mathtt{x^{2}} and constant.
\mathtt{\Longrightarrow \ 7\ \times 8}\\\ \\ \mathtt{\Longrightarrow \ 56}
(c) Find the coefficient of x
Coefficient of x is -15
(d) Find the value of a & b such that its multiplication produce 56 and addition produce -15.
This can be found by finding factor of 56.
Factor of 56 = 1, 2, 4, 7, 8, 14, 28 and 56.
Choose the two factor which will satisfy other condition.
The values of a & b are -8 & -7.
Checking the values.
-8 x -7 = 56
-8 + (-7) = -15
(e) Expand the middle value of quadratic expression using a & b and then factorize.
\mathtt{\Longrightarrow \ 7x^{2} -15x+8}\\\ \\ \mathtt{\Longrightarrow \ 7x^{2} -7x-8x+8}\\\ \\ \mathtt{\Longrightarrow \ 7x\ ( x-1) -8( x-1)}\\\ \\ \mathtt{\Longrightarrow \ ( x-1)( 7x-8)}
Factorizing Quadratic Polynomial – Practice Problems
Factorize the below expressions;
\mathtt{( i) \ x^{2} +12x+20}\\\ \\ \mathtt{( ii) \ x^{2} -18x+56}\\\ \\ \mathtt{( iii) \ x^{2} +x-2}\\\ \\ \mathtt{( iv) \ 3x^{2} +x+4}
Solution
\mathtt{( i)( x+2)( x+10)}\\\ \\ \mathtt{( ii) \ ( x-4)( x-14)}\\\ \\ \mathtt{( iii) \ ( x-1)( x+2)}\\\ \\ \mathtt{( iv) \ ( 3x+4)( x-1)}
Next chapter : Factorizing expression \mathtt{a^{3} -b^{3}}