Are you looking for the different problems on Factorization by Grouping? Then, you are in the right place. We have given all types of factorization problems on our website. Students can learn to Factorize by Grouping the Terms in these articles. While solving Factorization by grouping problems, students need to group the terms with common factors before factoring.
How to Factor by Grouping?
Have a look at the Factoring by Grouping Steps and learn how to solve related problems easily. Follow the guidelines provided and perform factorization by grouping method. They are as under
- Note down the given expression, group the first two terms and last two terms.
- Find out the greatest common factor(GCF) from the first term and second term.
- Now, find the common factor from the above two groups.
- Finally factor out the terms in terms of product.
Factorization by Grouping Examples
1. Factor grouping the expressions?
1 + x + xy + x²y.
Solution: Given Expression is 1 + x + xy + x²y.
Group the first two terms and last two terms.
First two terms are 1 + x and the last two terms are xy + x²y.
(1+ x) + (xy + x²y).
Find out the common factor from the above two groups.
(1+x) + xy(1+x).
Factor out the terms in terms of product.
(1+x) (1+xy).
By factor grouping the expression 1 + x + xy + x²y., we will get (1+x) (1+xy).
2. How to factor by grouping the following algebraic expressions?
(i) x² – xy + xz – zy.
Solution:
Given Expression is x² – xy + xz – zy
Group the first two terms and last two terms.
First two terms are x² – xy and the last two terms are xz – zy.
(x² – xy) + (xz – zy)
Find out the common factor from the above two groups.
x(x – y) + z(x – y)
Factor out the terms in terms of product.
(x + z) (x – y)
Factor by grouping the expression x² – xy + xz – zy, we will get the result as (x+z) (x-y).
(ii) x² + 3x + xy + 3y.
Solution:
Given Expression is x² + 3x + xy + 3y.
Group the first two terms and last two terms.
First two terms are x² + 3x and the last two terms are xy + 3y.
(x² + 3x) + (xy + 3y)
Find out the common factor from the above two groups.
x(x + 3) + y(x + 3)
Factor out the terms in terms of product.
(x+y) (x+3).
Factor by grouping the expression x² + 3x + xy + 3y, we will get solution as (x+y) (x+3).
3. Factorize the algebraic expressions.
(i) 2a + ba + 2b + b²
Solution:
Given Expression is 2a + ba + 2b + b²
Group the first two terms and last two terms.
First two terms are 2a + ba and the last two terms are 2b + b².
(2a + ba)+ (2b + b²).
Find out the common factor from the above two groups.
a(2 + b) + b(2 + b).
Factor out the terms in terms of product.
(a + b) (2 + b).
By factorizing the expression 2a + ba + 2b + b², we will get (a + b) (2 + b).
(ii) b² – yb + 5b– 5y.
Solution:
Given Expression is b² – yb + 5b– 5y.
Group the first two terms and last two terms.
First, two terms are b² – yb and the last two terms are 5b – 5y.
(b² – yb)+ (5b – 5y).
Find out the common factor from the above two groups.
b(b – y) + 5(b – y).
Factor out the terms in terms of product.
(5 + b) (b – y).
By factorizing the expression b² – yb + 5b– 5y, we will get (5 + b) (b – y).
(iii) pq – rq – ps + rs.
Solution:
Given Expression is pq – rq – ps + rs.
Group the first two terms and last two terms.
First two terms are pq – rq and the last two terms are – ps + rs.
(pq – rq) – (ps – rs).
Find out the common factor from the above two groups.
q(p – r) -s(p – r).
Factor out the terms in terms of product.
(p – r) (q – s).
By factorizing the expression pq – rq – ps + rs, we will get (p – r) (q – s).
(iv) ab – 2ac – db + 2dc.
Solution:
The given expression is ab – 2ac – db + 2dc.
Group the first two terms and last two terms.
Fist two terms are ab – 2ac and the last two terms are – db + 2dc.
(ab – 2ac) – (db – 2dc).
Find out the common factor from the above two groups.
a(b – 2c) -d(b – 2c).
Factor out the terms in terms of product.
(b – 2c) (a – d).
By factorizing the expression ab – 2ac – db + 2dc., we will get (b – 2c) (a – d).
(v) pq² – 3rqs – pqs + 3rs²
Solution:
The given expression is pq² – 3rqs – pqs + 3rs²
Group the first two terms and last two terms.
First two terms are pq^2 – 3rqs and the last two terms are –pqs + 3rs².
(pq² – 3rqs) – (pqs – 3rs²).
Find out the common factor from the above two groups.
q(pq – 3rs) -s(pq – 3rs).
Factor out the terms in terms of product.
(pq – 3rs) (q – s).
By factorizing the expression pq² – 3rqs – pqs + 3rs², we will get (pq – 3rs) (q – s).
4. Factor each of the following expressions by grouping
(i) a² – 3a – ab + 3b.
Solution:
Given expression is a² – 3a – ab + 3b
Group the first two terms and last two terms.
First two terms are a² – 3a and the last two terms are – ab + 3b.
(a² – 3a) – (ab – 3b).
Find out the common factor from the above two groups.
a(a – 3) – b(a – 3).
Factor out the terms in terms of product.
(a – 3) (a – b).
By factorizing the expression a² – 3a – ab + 3b, we will get (a – 3) (a – b).
(ii) pq² + rq² + 2p + 2r.
Solution:
Given expression is pq² + rq² + 2p + 2r
Group the first two terms and last two terms.
First two terms are pq² + rq² and the last two terms are 2p + 2r.
(pq² + rq²) + (2p + 2r).
Find out the common factor from the above two groups.
q²(p +r) + 2(p+r).
Factor out the terms in terms of product.
(q²+ 2) (p + r).
By factorizing the expression pq² + rq² + 2p + 2r, we will get (q²+ 2) (p + r).
(iii) 2pq² + 3pqr – 2sqr – 3sr²
Solution:
Given expression is 2pq² + 3pqr – 2sqr – 3sr²
Group the first two terms and last two terms.
First two terms are 2pq² + 3pqr and the last two terms are – 2sqr – 3sr².
(2pq² + 3pqr) – ( 2sqr + 3sr²).
Find out the common factor from the above two groups.
pq(2q + 3r) – sr(2q + 3r).
Factor out the terms in terms of product.
(2q + 3r) (pq – sr).
By factorizing the expression 2pq² + 3pqr – 2sqr – 3sr², we will get (2q + 3r) (pq – sr).
(iv) par² + qars – pnrs – qns²
Solution:
The given expression is par² + qars – pnrs – qns²
Group the first two terms and last two terms.
First two terms are par² + qars and the last two terms are – pnrs – qns².
(par² + qars) – (pnrs – qns²).
Find out the common factor from the above two groups.
ar(pr + qs) – ns(pr + qs).
Factor out the terms in terms of product.
(pr + qs) (ar – ns).
By factorizing the expression par² + qars – pnrs – qns², we will get (pr + qs) (ar – ns).
5. Factorize
(i) (a + b) (2a + 5) – (a + b) (a + 3).
Solution:
The given expression is (a + b) (2a + 5) – (a + b) (a + 3).
Find out the common factor from the above expression.
(a + b) [(2a + 5) – (a + 3)].
Expand the terms.
(a + b)(2a + 5 – a – 3).
Simplify the second term.
(a + b) (a + 2).
By factorizing the expression (a + b) (2a + 5) – (a + b) (a + 3), we will get (a + b)(a + 2).
(ii) 6xy – y² + 12xz – 2yz.
Solution:
The given expression is 6xy – y² + 12xz – 2yz.
Group the first two terms and last two terms.
First two terms are 6xy – y² and the last two terms are 12xz – 2yz.
(6xy – y²) + (12xz – 2yz).
Find out the common factor from the above two groups.
y(6x – y)+ 2z(6x – y).
Factor out the terms in terms of product.
(6x – y) (y + 2z).
By factorizing the expression 6xy – y² + 12xz – 2yz, we will get (6x – y) (y + 2z).