Grade 11 Maths Questions With Detailed Solutions

The solutions and explanations to the questions on how to factor polynomials by grouping are presented.

## 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
## Question a)Factor 2 x^{ 2} - 4 x + x y - 2 y Solution a) We first find a common factor in 2 x and factor it as follows:^{ 2} - 4 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)## Question b)Factor x^{ 2} + 3 x - 2 x - 6Solution 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.
## Question c)Factor 15 x^{ 2} - 3 x + 10 x - 2Solution 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)## Question d)Factor4 x^{ 2} + x - 3Solution 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)
## Question e)Factorx^{ 2} y + 3 x + x^{ 2} y^{ 2} + 3 x ySolution 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)## Question f)Factor3 x^{ 2} + 3 x y - x + 2 y - 2Solution 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) |

High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers

Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers

Home Page