# Factorize by Grouping the Terms | How to Factor by Grouping Terms?

Factorize by grouping the terms consists of two or more product terms as resultant value. To divide the expression into product terms, we need to identify the greatest common factor which divides all the remaining terms in the expression. Step by step process on how to do Factorization by Grouping the Terms are clearly given in this article.

## Procedure to Factorize by Grouping the Terms

Refer to the below-mentioned step by step process and learn the Factoring by Grouping Terms. They are along the lines

1. Note down the given expression.
2. Group the first two terms and last two terms.
3. Factor out the greatest common factor from each group.
4. Finally, you will get two or more product terms as result.

### Solved Examples on Factorization by Grouping Terms

1. Factorize grouping the following expressions?

(i) 18x³y³ – 27x²y³ + 36x³y²

Solution:
The given expression is 18x³y³ – 27x²y³ + 36x³y²
Factor out the greatest common factor from the above expression.
That is, 9x²y²(2xy – 3y + 4x).

Therefore, the final solution for the expression 18x³y³ – 27x²y³ + 36x³y² is 9x²y²(2xy – 3y + 4x).

(ii) 12a²b³ – 21a³b²

Solution:
The given expression is 12a²b³ – 21a³b²
Factor out the greatest common factor from the above expression.
3a²b²(4b – 7a).

Finally, the solution for the expression 12a²b³ – 21a³b² is 3a²b²(4b – 7a).

(iii) a³ – a² + a – 1.

Solution:
The given expression is a³ – a² + a – 1
Group the first two terms and last two terms.
Here, first two terms are a³ – a² and the last two terms are a – 1.
So, (a³ – a²) + (a – 1).
Now, factor out the greatest common factor from each group.
That is, a²(a – 1) + (a – 1).
(a – 1) (a² + 1).

Therefore, solution for the expression a³ – a² + a – 1 is (a – 1) (a² + 1).

(iv) prs + qurs – pt – qut

Solution:
The given expression is prs + qurs – pt – qut.
Group the first two terms and last two terms.
Here, the first two terms are prs + qurs and last two terms are – pt – qut.
Then, (prs + qurs) – (pt + qut).
Now, factor out the greatest common factor from the above two groups.
That is, rs(p + qu) – t(p + qu).
(p + qu) (rs – t).

Therefore, the solution for the expression prs + qurs – pt – qut is (p + qu) (rs – t).

(v) a² – 3a – ab + 3b
Solution:

The given expression is
a² – 3a – ab + 3b.
Group the first two terms and last two terms.
Here, the first two terms are a²-3a and the last two terms are ab + 3b.
Then, (a² – 3a) – (ab – 3b).
Now, factor out the greatest common factor from the above two groups.
That is a(a – 3) – b(a – 3).
(a – 3) (a – b).

Therefore, solution for the expression is a² – 3a – ab + 3b is (a – 3) (a – b).

2. How to factorize by grouping the following expressions?
(i) 2q^4 – q³ + 4q – 2
Solution:
The given expression is
2q^4 – q³ + 4q – 2.
Group the first two terms and last two terms.
Here, the first two terms are 2q^4 – q³ and the last two terms are 4q – 2.
Then, (2q^4 – q^3) + (4q-2).
Now, factor out the greatest common factor from the above two groups.
That is, q^3(2q – 1) +2(2q – 1).
(2q – 1) (q^3 + 2).

Therefore, solution for the expression is 2q4 – q3 + 4q – 2 is (2q-1) (q^3+2).

(ii) ac + bc – ad – bd.

Solution:
The given expression is ac + bc – ad – bd.
Group the first two terms and last two terms.
Here, the first two terms are ac+bc and the last two terms are – ad-bd.
Then, (ac + bc) – (ad + bd).
Now, factor out the greatest common factor from the above two groups.
That is, c(a+b) -d(a+b).
(a+b) (c-d).

Therefore, solution for the expression is ac + bc – ad – bd is (a+b) (c-d).

(iii) pa – pb – qa –qb.

Solution:
The given expression is pa – pb – qa–qb.
Group the first two terms and last two terms.
Here, the first two terms are pa – pb, and the last two terms are qa – qb.
Then, (pa – pb) – (qa -qb).
Now, factor out the greatest common factor from the above two groups.
That is, p(a – b) – q(a – b).
(p – q) (a – b).

Therefore, the solution for the expression is pa – pb – qa – qb is (p – q) (a – b).

3. How to factorize by grouping the algebraic expressions?
(i)x^2z^2 + xzw + xyz + yw

Solution:
The given expression is
x^2z^2 + xzw + xyz + yw.
Group the first two terms and last two terms.
Here, the first two terms are x^2z^2 + xzw and the last two terms are xyz + yw.
Then, (x^2z^2 + xzw) + (xyz + yw).
Now, factor out the greatest common factor from the above two groups.
That is, xz(xz + w) + y(xz + w).
(xz + y) (xz + w).

Therefore, solution for the expression is x^2z^2 + xzw + xyz + yw is (xz + y) (xz + w).

(ii) 5x + xy + 5y + y^2

Solution:
The given expression is 5x + xy + 5y + y^2
Group the first two terms and last two terms.
Here, the first two terms are 5x + xy, and the last two terms are 5y + y^2.
Then, (5x + xy) + (5y + y^2).
Now, factor out the greatest common factor from the above two groups.
That is, x(5 + y) + y(5 + y).
(5 + y) (x + y).

Therefore, solution for the expression is 5x + xy + 5y + y^2 is (5 + y) (x + y).

(iii) xy – yz – xz + z^2

Solution:
The given expression is xy – yz – xz + z^2
Group the first two terms and last two terms.
Here, the first two terms are xy – yz and the last two terms are – xz + z^2.
Then, (xy – yz) – (xz – z^2).
Now, factor out the greatest common factor from the above two groups.
That is, y(x – z) – z(x – z).
(x – z) (y – z).

Therefore, solution for the expression xy – yz – xz + z^2 is (x – z) (y – z).

4. Factorize the expressions
(i) a^4 + a^3 + 2a + 2

Solution:
The given expression is a^4 + a^3 + 2a + 2.
Group the first two terms and last two terms.
Here, the first two terms are a^4 + a^3 and the last two terms are 2a + 2.
Then, (a^4 + a^3) + (2a + 2).
Now, factor out the greatest common factor from the above two groups.
That is, a^3(a + 1) + 2(a + 1).
(a + 1) (a^3 + 2).

Therefore, solution for the expression a^4 + a^3 + 2a + 2 is (a + 1) (a^3 + 2).

(ii) a²b² + d²b² – cd² – ca²

Solution:
The given expression is a²b² + d²b² – cd² – ca²
Group the first two terms and last two terms.
Here, the first two terms are a²b² + d²b² and the last two terms are – cd² – ca².
Then, (a^2b^2 + d^2b^2) – (cd^2 + ca^2).
Now, factor out the greatest common factor from the above two groups.
That is, b^2(a^2 + d^2) – c(d^2 + a^2).
(a^2 + d^2) (b^2 – c).

Therefore, solution for the expression a2b2 + d2b2 – cd2 – ca2 is(a^2 + d^2) (b^2 – c).

5. Factorize by grouping the terms
(x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y

Solution:
Thegiven expression is (x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y.
Group the first two terms and last two terms.
here, first two terms are (x^2 + 3x)^2 – 2(x^2 + 3x) and the last two terms are –y(x^2 + 3x) + 2y.
Then, [(x^2 + 3x)^2 – 2(x^2 + 3x)] –[y(x^2 + 3x) – 2y].
That is, (x^2 + 3x)(x^2 + 3x – 2) – y(x^2 + 3x – 2).
(x^2 + 3x – 2)(x^2 + 3x – y).

Therefore, solution for the expression (x2 + 3x)2 – 2(x2 + 3x) – y(x2 + 3x) + 2y is (x^2 + 3x – 2)(x^2 + 3x – y).