# Factor Polynomials by Grouping

Questions With Solutions

How to factor a polynomial by grouping? Questions with detailed solutions and explanations are presented.

## Questions with Solutions

Question 1

Factor completely the polynomial *4 x ^{ 2} + 4 x + 3 x + 3*

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)
Question 2

Factor the polynomial *2 x ^{ 2} - 4 x + 3 x y - 6 y*

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}- 4x + 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)
Question 3

Factor the polynomial *x y - x - 2 y + 2*

Solution to Question 3

The terms in the given polynomial have no common factor.

The first two terms can be grouped and factored as follows: *x* out:

*x y - x = x ( y - 1)*

The last two terms can factored as: *2* out:

*- 2 y + 2 = 2( - y + 1) = - 2(y - 1)*

Rewrite the given polynomial in factored form as follows:

*x y - x - 2 y + 2 = x (y - 1) - 2 (y - 1) *

Factor out the common factor *(y - 1)* to factor completely

*x y - x - 2 y + 2 = (y - 1) - 2 (y - 1) = (y - 1)(x - 2)*

Question 4

Factor completely the polynomial 3* x ^{ 2} + 4 x + 1*

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 + 1*3 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 completely

a)*2 x*

^{ 2}- 4 x + x y - 2 yb)

*x*

^{ 2}+ 3 x - 2 x - 6c)

*15 x*

^{ 2}- 3 x + 10 x - 2d)

*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)

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)*

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 Variable__

The 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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