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
Question 2
Question 3
Question 4
More Questions on Factoring Polynomials
Use grouping to factor the following polynomials completelya) 2 x^{ 2} 4 x + x y  2 yb) 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)
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) 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 xintercepts 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) 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) 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) ) 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
