# Factor Polynomials by Grouping

Questions With Detailed 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
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)
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)
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 + 13 x
^{ 2} + 3 x + x + 1We 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 completelya) 2 x^{ 2}- 4 x + x y - 2 y b) x^{ 2} + 3 x - 2 x - 6c) 15 x^{ 2} - 3 x + 10 x - 2 d) 4 x^{ 2} + x - 3e) x^{ 2} y + 3 x + x^{ 2} y^{ 2} + 3 x yf) 3 x^{ 2} + 3 x y - x + 2 y - 2## Solutions to Above Questions
Solution to Question a)
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 and factor it as follows:^{ 2} + 3 x 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 VariableThe graph of the given polynomial in b) above y = x is shown below. ^{ 2} + 3 x - 2 x - 6 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 and factor it as follows:^{ 2} - 3 x 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 - 3We can now factor 4 x as follows:^{ 2} + 4 x 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 has the factor ^{ 2})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 - 2We shall now factor the equivalent polynomial 3 x. We can now group the first 3 terms and factor as follows:^{ 2} + 3 x y - 3 x + 2 x + 2 y - 2 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) |

### More References and links

Introduction to PolynomialsFactor Polynomials

Factoring of Special Polynomials

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