Factorize by Grouping The Terms

Factorize by grouping the terms (two or more) means that we need to group the terms which have common factors before factoring. 

Method to factorize by grouping the terms:

(i) From the groups of the given expression a common factor can be taken out from each group.

(ii) Factorize each group

(iii) Now take out the factor common to group formed.

Now we will learn how to factorize by grouping two or more terms.

Solved examples to factorize by grouping the terms:

1. Factorize grouping the following expressions:

(i) 18a³b³ - 27a²b3 + 36a3b²

Solution:

18a³b³ - 27a²b3 + 36a3b²

= 9a²b²(2ab – 3b + 4a)


(ii) 12x²y³ - 21x³y²

Solution:

12x²y³ - 21x³y²

= 3x²y²(4y - 7x)


(iii) y³ - y² + y - 1

Solution:

y³ - y² + y - 1

= y²(y - 1) + 1(y - 1)

= (y - 1) (y² + 1)


(iv) axy + bcxy – az – bcz

Solution:

axy + bcxy – az – bcz

= xy(a + bc) – z(a + bc)

= (a + bc) (xy – z)


(v) x² - 3x – xy + 3y

Solution:

x² - 3x – xy + 3y

= x(x - 3) - y(x - 3) 

= (x - 3) (x – y) 

2. How to factorize by grouping the following expressions?

(i) 2x⁴ – x³ + 4x - 2

Solution:

2x⁴ – x³ + 4x - 2

= x³(2x – 1) + 2(2x - 1)

= (2x – 1) (x³ + 2)

(ii) pr + qr - ps - qs

Solution: 

  pr + qr - ps - qs

= r(p + q) - s(p + q)

= (p + q) (r - s)

(iii) mx - my - nx - ny

Solution: 

mx - my - nx - ny

= m(x - y) - n(x - y)

= (x - y) (m - n)

 

3. How to factorize by grouping the algebraic expressions?

(i) a²c² + acd + abc + bd

Solution:

a²c² + acd + abc + bd

= ac(ac + d) + b(ac + d)

= (ac + d) (ac + b)


(ii) 5a + ab + 5b + b²

Solution:

5a + ab + 5b + b²

= a(5 + b) + b(5 + b)

= (5 + b) (a + b)


(iii) ab - by - ay + y²

Solution:

ab - by - ay + y²

= b(a - y) - y(a - y)

= (a - y) (b - y)

4. Factorize the expressions:

(i) x⁴ + x³ + 2x + 2

Solution:

x⁴ + x³ + 2x + 2

= x³(x + 1) + 2 (x + 1)

= (x + 1) (x³ + 2)


(ii) f²x² + g²x² – ag² – af²

Solution:

f²x² + g²x² – ag² – af²

= x²(f² + g²) – a (g² + f²)

= x²(f² + g²) – a(f² + g²)

= (f² + g²)(x² – a)



5. Factorize by grouping the terms (a² + 3a)² - 2(a² + 3a) – b(a² + 3a) + 2b

Solution:

(a² + 3a)² - 2(a² + 3a) – b(a² + 3a) + 2b

= [(a² + 3a)² - 2(a² + 3a)] – [b(a² + 3a) - 2b]

= (a² + 3a) (a² + 3a - 2) – b(a² + 3a - 2)

= (a² + 3a - 2) (a² + 3a - b)

8th Grade Math Practice

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