Solution to Factoring Polynomials by Grouping
Grade 11 Math 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 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


Question a): Factor 2 x 2 - 4 x + x y - 2 y

Solution

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)



Question b): Factor x 2 + 3 x - 2 x - 6

Solution

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.

graph of polynomial for solution in b) .


Conlusion: 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 - 2

Solution

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)



Question d): Factor 4 x 2 + x - 3

Solution

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)



Question e): Factor x 2 y + 3 x + x 2 y 2 + 3 x y

Solution

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)



Question f): Factor 3 x 2 + 3 x y - x + 2 y - 2

Solution

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)



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Updated: 20 January 2017 (A Dendane)





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