FACTOR BY GROUPING
Factor by grouping :
Whenever we have two or three terms, then we have to find the highest common factor for those terms then we have to factor out the common term.
In case we have four terms, then we have write first two terms together then write last two terms together.
After grouping the given terms, we have to find the highest common factor for first two terms and factor out, then find H.C.F for last two terms and factor out.
Factor by grouping - Examples
Example 1 :
Factorize x² + x y
Solution :
In the given question, we have two terms x²and x y.
H.C.F (x²and x y) = x
Now we are going to factor out "x" from both terms,
= x (x + y)
Example 2 :
Factorize 4a - 8b + 5ax - 10bx
Solution :
In the given question, we have four terms. So we have to consider the first two terms as one group and last two terms as one group.
H.C.F (4a and -8b) = 4
H.C.F (5ax and -10bx) = 5x
Now we are going to factor out "4" from first two terms, and factor out "5x" from last two terms
= 4 (a - 2b) + 5x (a - 2b)
= (4 + 5x) (a - 2b)
Example 3 :
Factorize pq + pr - 3ps
Solution :
In the given question, we have three terms. So we have to find the H.C.F of pq, pr and -3ps.
H.C.F (pq, pr and -3ps) = p
Now we are going to factor out "p" from all terms.
= p (q + r - 3s)
Example 4 :
Factorize 2 a³ + 4 a²
Solution :
In the given question, we have two terms. So we have to find the H.C.F of 2 a³ and 4 a²
H.C.F (2 a³, 4 a²) = 2 a²
Now we are going to factor out "2a²" from all terms.
= 2a² (a + 2)
Example 5 :
Factorize 6 a⁵ - 18 a³ + 42 a²
Solution :
In the given question, we have three terms. So we have to find the H.C.F of 6 a⁵, - 18 a³, 42 a²
H.C.F (6 a⁵, - 18 a³, 42 a²) = 6 a²
Now we are going to factor out "6 a²" from all terms.
= 6 a² (a³ - 3a + 7)
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Related topics
- Factoring quadratic equations when the coefficient of x² is not 1
- Solving quadratic equation using sompleting the square method
- Solving quadratic equation by using quadratic formula
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