# Factorization of polynomial by grouping

In this chapter we will learn to factorize different polynomials by grouping.

Here we will state the exact method along with solved problems so that you can understand the concept with full clarity.

## Factorization by grouping

Follow the below step to factorize the polynomial;

(a) Divide the polynomial into different groups.

(b) Factorize each group individually.

(c) Now find the common elements present in all the group.

(d) Separate the common element and then simplify further.

In the above 4 steps you can factorize the polynomial into individual components.

Note:
This method doesn’t work in all the cases.
With practice, you will learn to identify the problem and figure out if the grouping method will work or not.

Let us see some solved examples for further clarity.

Example 01
Factorize the polynomial by grouping method
\mathtt{\Longrightarrow \ 2+5a\ +\ 8ab\ +\ 20a^{2} b}

Solution
Form group of first and last two entities.

\mathtt{\Longrightarrow \ 2+5a\ +\ 8ab\ +\ 20a^{2} b}\\\ \\ \mathtt{\Longrightarrow \ ( 2\ +\ 5a) \ +\left( 8ab\ +\ 20a^{2} b\right)}

Now factorize each of the group individually.

\mathtt{\Longrightarrow \ ( 2\ +\ 5a) \ +\left( 8ab\ +\ 20a^{2} b\right)}\\\ \\ \mathtt{\Longrightarrow \ ( 2+5a) \ +\ 4ab( 2+5a)}

Note that (2 + 5a) binomial is common.

Simplifying further using common binomial method;

\mathtt{\Longrightarrow \ ( 2+5a) \ +\ 4ab( 2+5a)}\\\ \\ \mathtt{\Longrightarrow \ ( 2+5a) \ ( 1+4ab)}

Hence, the given polynomial is fully factorized to (2+5a).(1+4ab)

Example 02
Factorize using grouping method.
\mathtt{\Longrightarrow \ 7x^{2} +5x\ +21x\ +15}

Solution
\mathtt{\Longrightarrow \ 7x^{2} +5x\ +21x\ +15}\\\ \\ \mathtt{\Longrightarrow \ \left( 7x^{2} +5x\right) \ +( 21x\ +15)}\\\ \\ \mathtt{\Longrightarrow \ x\ ( 7x+5) \ +\ 3\ ( 7x+5)}\\\ \\ \mathtt{\Longrightarrow \ ( 7x+5) \ ( x+3)}

Hence, we got the factors ( 7x + 5 ) ( x + 3 ) after factorization.

Example 03
Factorize the polynomial
\mathtt{\Longrightarrow \ 4x^{2} +5x\ +\ 4}

Solution
In the polynomial, we have only three terms.

To form group of two’s we need at-least four elements.

Break 5x into two parts such that we can have groups with common elements.

\mathtt{\Longrightarrow \ 4x^{2} +5x\ +\ 4}\\\ \\ \mathtt{\Longrightarrow \ 4x^{2} +4x\ +x\ +1}\\\ \\ \mathtt{\Longrightarrow \ \left( 4x^{2} +4x\right) \ +\ ( x\ +1)}\\\ \\ \mathtt{\Longrightarrow \ 4x\ ( x+1) \ +\ ( x+1)}\\\ \\ \mathtt{\Longrightarrow \ ( x+1) \ ( 4x+1)}

Hence, on factorization we get (x + 1) (4x + 1)

Example 04
Factorize the below polynomial by grouping method.
\mathtt{\Longrightarrow \ 2x^{3} +3y^{3} +2xy^{2} +3x^{2} y} [/latex]

Solution
Arrange the entities such that we can form group which can be factorized easily.

\mathtt{\Longrightarrow \ 2x^{3} +3y^{3} +2xy^{2} +3x^{2} y}\\\ \\ \mathtt{\Longrightarrow \ \ 2x^{3} +2xy^{2} +3y^{3} +3x^{2} y}

Now form group of first and last two terms.

\mathtt{\Longrightarrow \left( 2x^{3} +2xy^{2}\right) +\left( 3y^{3} +3x^{2} y\right)}\\\ \\ \mathtt{\Longrightarrow \ 2x\left( x^{2} +y^{2}\right) +3y\left( y^{2} +x^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ 2x\left( x^{2} +y^{2}\right) +3y\left( x^{2} +y^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} +y^{2}\right)( 2x+3y)}

Example 05
\mathtt{\Longrightarrow \ 5x\ +\ bx\ +\ 5b\ +\ b^{2}}

Solution
\mathtt{\Longrightarrow \ 5x\ +\ bx\ +\ 5b\ +\ b^{2}}\\\ \\ \mathtt{\Longrightarrow \ ( 5x\ +\ bx) \ +\ \left( 5b\ +\ b^{2}\right)}\\\ \\ \mathtt{\Longrightarrow \ x\ ( 5+b) \ +\ b\ ( 5+b)}\\\ \\ \mathtt{\Longrightarrow \ ( 5+b) \ ( x+b)}

Example 06
Factorize by grouping method
\mathtt{\Longrightarrow \ 5x^{3} \ -\ 10x^{2} \ +\ 7x \ -14\ }

Solution
\mathtt{\Longrightarrow \ 5x^{3} \ -\ 10x^{2} \ +\ 7x \ -14\ }\\\ \\ \mathtt{\Longrightarrow \left( 5x^{3} \ -\ 10x^{2}\right) \ +\ ( 7x-14)}\\\ \\ \mathtt{\Longrightarrow \ 5x^{2} \ ( x-2) \ +\ 7\ ( x-2)}\\\ \\ \mathtt{\Longrightarrow \ ( x-2) \ \left( 5x^{2} +7\right)}

Example 07
Factorize using grouping method
\mathtt{\Longrightarrow \ x^{2} -4x-12}

Solution
Here the polynomial contain three entities.

To form group of two’s, we need at least four entities.

Let’s break -4x into two parts.

Now -4x can be broken into multiple ways like;

⟹ -x – 3x

⟹ -2x – 2x

⟹ -6x + 2x

You have to choose the one which gives common element upon factorization.

Let’s take the first option -4x ⟹ (- x – 3x )

\mathtt{\Longrightarrow \ x^{2} -4x-12}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -x-3x-12}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} -x\right) -( 3x+12)}\\\ \\ \mathtt{\Longrightarrow \ x( x-1) -3\ ( x+4)}

After this point we cannot simplify the expression further.

Hence, -4x ⟹ (- x – 3x ) is the wrong choice for factorization.

Similar is the case for second option, -4x ⟹ -2x – 2x

Now let’s take the third option, -4x ⟹ -6x + 2x

\mathtt{\Longrightarrow \ x^{2} -4x-12}\\\ \\ \mathtt{\Longrightarrow \ x^{2} -6x+2x-12}\\\ \\ \mathtt{\Longrightarrow \ \left( x^{2} -6x\right) +( 2x-12)}\\\ \\ \mathtt{\Longrightarrow \ x( x-6) +2\ ( x-6)}\\\ \\ \mathtt{\Longrightarrow \ ( x-6) \ ( x+2)}

Hence, the polynomial is factorized to (x – 6) ( x + 2)

Note:
Here breaking the middle term into right component may consume lot of time. This type of polynomial can be easily solved using quadratic equations.

Next chapter : Factorization of polynomials using formulas

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