# Factor Polynomials by Grouping Questions With Solutions

How to factor a polynomial by grouping? Questions with detailed solutions and explanations are presented.

Factoring a polynomial by grouping is explained using several questions with their solutions.

## Questions with Solutions

Question 1
Factor completely the polynomial 4 x 2 + 4 x + 3 x + 3
Solution to Question 1
Note that all four terms in the given polynomial have no common factor.
However by grouping the first two term, we can factor 4 x out as follows:
4 x 2 + 4 x = 4 x (x + 1)
We now group the last two term and factor 3 out as follows:
3 x + 3 = 3 ( x + 1)
Rewrite the given polynomial with the grouped terms in factored form.
4 x 2 + 4 x + 3 x + 3 = 4 x (x + 1) + 3 ( x + 1)
Note that ( x + 1) is a common factor which can be factored out as follows:
4 x 2 + 4 x + 3 x + 3 = 4 x (x + 1) + 3 ( x + 1) = (x + 1)(4x + 3)

Question 2
Factor the polynomial 2 x 2 - 4 x + 3 x y - 6 y
Solution to Question 2
There is no common factor to all 4 terms in the given polynomial.
Group the first two terms and factor 2 x out :
2 x 2 - 4 x = 2 x ( x - 2)
Group the last two terms and factor 3 y out :
3 x y - 6 y = 3 y( x - 2)
Rewrite the given polynomial as follows
2 x 2 + 2 x + 3 x y + 3 y = 2 x (x - 2) + 3 y(x - 2)
and factor out the common factor (x - 2).
2 x 2 + 2 x + 3 x y + 3 y = 2 x ( x - 2) + 3 y( x - 2) = (x - 2)(2 x + 3 y)

Question 3
Factor the polynomial x y - x - 2 y + 2
Solution to Question 3
The terms in the given polynomial have no common factor.
The first two terms can be grouped and factored as follows: x out:
x y - x = x ( y - 1)
The last two terms can factored as: 2 out:
- 2 y + 2 = 2( - y + 1) = - 2(y - 1)
Rewrite the given polynomial in factored form as follows:
x y - x - 2 y + 2 = x (y - 1) - 2 (y - 1)
Factor out the common factor (y - 1) to factor completely
x y - x - 2 y + 2 = (y - 1) - 2 (y - 1) = (y - 1)(x - 2)

Question 4
Factor completely the polynomial 3 x 2 + 4 x + 1
Solution to Question 4
There is no common factor to the terms in the given polynomial. One way is to rewrite the polynomial with 4 terms that may be factored by grouping.
We use the identity 4 x = 3 x + x to rewrite the given polynomial as follows:
3 x 2 + 4 x + 1 = 3 x 2 + 3 x + x + 1
We group the first two terms and factor 3 x out as follows:
3 x 2 + 3 x = 3 x (x + 1)
Rewrite the given polynomial with the grouped terms in factored form.
3 x 2 + 4 x + 1 = 3 x 2 + 3 x + x + 1 = 3 x (x + 1) + 1( x + 1)
Note that ( x + 1) is a common factor which can be factored out as follows:
3 x 2 + 4 x + 1 = 3 x (x + 1) + 1( x + 1) = (x + 1)(3 x + 1)

## More Questions on Factoring Polynomials

### Use grouping to factor the following polynomials completely

a) 2 x 2- 4 x + x y - 2 y
b) x 2 + 3 x - 2 x - 6
c) 15 x 2 - 3 x + 10 x - 2
d) 4 x 2 + x - 3
e) x 2 y + 3 x + x 2 y 2 + 3 x y
f) 3 x 2 + 3 x y - x + 2 y - 2

## Solutions to Above Questions

Solution to Question a)
We first find a common factor in 2 x 2 - 4 and factor it as follows:
2 x 2 - 4 x = 2 x (x - 2)
We next find a common factor in the x y - 2 y and factor it as follows:
x y - 2 y = y (x - 2)
Use the common factor (x - 2) and factor the given polynomial as follows:
2 x 2 - 4 x + x y - 2 y = ( 2 x 2 - 4 x ) + (x y - 2 y)

= 2 x (x - 2) + y (x - 2) = (x - 2)(2x + y)

Solution to Question b)
Find a common factor in x 2 + 3 x and factor it as follows:
x 2 + 3 x = x (x + 3)
We next find a common factor in the - 2 x - 6 and factor it as follows:
- 2 x - 6 = - 2 (x + 3)
Use the common factor (x + 3) to factor the given polynomial as follows:
x 2 + 3 x - 2 x - 6 = ( x 2 + 3 x ) + (- 2 x - 6)
= x (x + 3) - 2 (x + 3) = (x + 3)(x - 2)

Graphical Interpretation of Factorization for a Polynomial in one Variable
The graph of the given polynomial in b) above y = x 2 + 3 x - 2 x - 6 is shown below. x = - 3 makes the factor (x + 3) equal to zero and x = 2 makes the factor x - 2 equal to zero. Both x = -3 and x = 2 appears as x-intercepts in the graph of the given polynomial.

Conclusion: One way to check our factoring is to graph the given polynomial and check that the x intercepts corresponds to the zeros of the factors included in the factorization.

Solution to Question c)
Find a common factor in 15 x 2 - 3 x and factor it as follows:
15 x 2 - 3 x = 3 x (5 x - 1)
We next find a common factor in the 10 x - 2 and factor it as follows:
10 x - 2 = 2 (5 x - 1)
Use the common factor (5 x - 1) to factor the given polynomial as follows:
15 x 2 - 3 x + 10 x - 2 = ( 15 x 2 - 3 x ) + (10 x - 2)
= 3 x (5 x - 1) + 2 (5 x - 1) = (5 x - 1)(3 x + 2)

Solution to Question d)
The given polynomial has three terms with no common factor. One way to factor is to rewrite it replacing x by 4 x - 3 x as follows:
4 x 2 + x - 3 = 4 x 2 + 4 x - 3 x - 3
We can now factor 4 x 2 + 4 x as follows:
4 x 2 + 4 x = 4 x (x + 1)
We next factor - 3 x - 3 as follows:
- 3 x - 3 = - 3 (x + 1)
Use the common factor (x + 1) to factor the given polynomial as follows:
4 x 2 + x - 3 = 4 x 2 + 4 x - 3 x - 3 = (4 x 2 + 4 x) + (- 3 x - 3)
= 4 x (x + 1) - 3 (x + 1) = (x + 1)(4 x - 3)

Solution to Question e)
Note that x is a common factor to all terms in the given polynomial. Hence we start by factoring as follows:
x 2 y + 3 x + x 2 y 2 + 3 x y = x( x y + 3 + x y 2 + 3 y)
Rewrite by grouping terms as follows
x 2 y + 3 x + x 2 y 2 + 3 x y = x( (x y + x y 2) + (3 + 3 y) )
The terms in (x y + x y 2) has the factor x y and the terms in (3 + 3 y) has the common factor 3. Hence we factor as follows
x 2 y + 3 x + x 2 y 2 + 3 x y = x( (x y + x y 2) + (3 + 3 y) )
= x( x y (1 + y) + 3 (1 + y) ) = x (1 + y)( x y + 3)

Solution to Question f)
Note that there are 5 terms in the given polynomial with common factor to all of them. Rewrite the polynomial replacing - x by - 3 x + 2 x as follows.
3 x 2 + 3 x y - x + 2 y - 2 = 3 x 2 + 3 x y - 3 x + 2 x + 2 y - 2
We shall now factor the equivalent polynomial 3 x 2 + 3 x y - 3 x + 2 x + 2 y - 2. We can now group the first 3 terms and factor as follows:
3 x 2 + 3 x y - 3 x = 3 x (x + y - 1)
We now group the last three terms and factor as follows
2 x + 2 y - 2 = 2 (x + y - 1 )
The two groups have the common factor (x + y - 1 ) and the given polynomial is factored as follows:
3 x 2 + 3 x y - x + 2 y - 2 = 3 x 2 + 3 x y - 3 x + 2 x + 2 y - 2
= (3 x 2 + 3 x y - 3 x) + (2 x + 2 y - 2)
= 3x ( x + y - 1) + 2(x + y - 1) = (x + y - 1)(3x + 2)