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.

.

__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