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# Factor by Grouping – Methods & Examples

Now that you have learned how to factor polynomials by using different methods such as; Greatest Common Factor (GCF, Sum or difference in two cubes; Difference in two squares method; and Trinomial method.

*Which method do you find simplest among these? *

All these methods of factoring polynomials are as easy as ABC, only if they are applied correctly.

**In this article, we will learn another simplest method known as factoring by Grouping, but before getting into this topic of factoring by grouping, let’s discuss what factoring a polynomial is.**

A polynomial is an algebraic expression with one or more terms in which an addition or a subtraction sign separates a constant and a variable.

The general form of a polynomial is ax^{n} + bx^{n-1 }+ cx^{n-2 }+ …. + kx + l, where each variable has a constant accompanying it as its coefficient. The different types of polynomials include; binomials, trinomials, and quadrinomial.

Examples of polynomials are; 12x + 15, 6x^{2} + 3xy – 2ax – ay, 6x^{2} + 3x + 20x + 10 etc.

## How to Factor by Grouping?

**Factor by Grouping** is useful when there is no common factor among the terms, and you split the expression into two pairs and factor each of them separately.

**Factoring polynomials** is the reverse operation of multiplication because it expresses a polynomial product of two or more factors. You can factor polynomials to find the roots or solutions of an expression.

### How to factor trinomials by grouping?

To factor a trinomial of the form ax^{2} + bx + c by grouping, we carry out the procedure as shown below:

- Find the product of the leading coefficient “a” and the constant “c.”

⟹ a * c = ac

- Look for the factors of the “ac” that add to coefficient “b.”
- Rewrite bx as a sum or difference of the factors of ac that add to b.

⟹ ax^{2 }+ bx + c = ax^{2 }+ (a + c) x + c

⟹ ax^{2 }+ ax + cx + c

- Now factor by grouping.

⟹ ax (x + 1) + c (x + 1)

⟹ (ax + c) (x + 1)

*Example 1*

Factor x^{2 }– 15x + 50

__Solution__

Find the two numbers whose sum is -15 and product is 50.

⟹ (-5) + (-10) = -15

⟹ (-5) x (-10) = 50

Rewrite the given polynomial as;

x^{2}-15x + 50⟹ x^{2}-5x – 10x + 50

Factorize each set of groups;

⟹ x(x – 5) – 10(x – 5)

⟹ (x – 5) (x – 10)

*Example 2*

Factor the trinomial 6y^{2} + 11y + 4 by grouping.

__Solution__

6y^{2} + 11y + 4 ⟹ 6y^{2} + 3y + y + 4

⟹ (6y^{2} + 3y) + (8y + 4)

⟹ 3y (2y + 1) + 4(2y + 1)

= (2y + 1) (3y + 4)

*Example 3*

Factor 2x^{2 }– 5x – 12.

__Solution__

2x^{2 }– 5x – 12

= 2x^{2 }+ 3x – 8x – 12

= x (2x + 3) – 4(2x + 3)

= (2x + 3) (x – 4)

*Example 4*

Factor 3y^{2} + 14y + 8

__Solution__

3y^{2} + 14y + 8 ⟹ 3y^{2} + 12y + 2y + 8

⟹ (3y^{2} + 12y) + (2y + 8)

= 3y (y + 4) + 2(y + 4)

Hence,

3y^{2} + 14y + 8 = (y + 4) (3y + 2)

*Example 5*

Factor 6x^{2}– 26x + 28

__Solution__

Multiply the leading coefficient by the last term.

⟹ 6 * 28 = 168

Find two numbers whose sum is product is 168 and sum is -26

⟹ -14 + -12 = -26 and -14 * -12 = 168

Write the expression by replacing bx with the two numbers.

⟹ 6x^{2}– 26x + 28 = 6x^{2} + -14x + -12x + 28

6x^{2} + -14x + -12x + 28 = (6x^{2} + -14x) + (-12x + 28)

= 2x (3x + -7) + -4(3x + -7)

Therefore, 6x^{2}– 26x + 28 = (3x -7) (2x – 4)

### How to factor binomials by grouping?

A binomial is an expression with two terms combined by either addition or subtraction sign. To factor a binomial, the following four rules are applied:

- ab + ac = a (b + c)
- a
^{2}– b^{2}= (a – b) (a + b) - a
^{3}– b^{3}= (a – b) (a^{2}+ab + b^{2}) - a
^{3}+ b^{3}= (a + b) (a^{2}– ab + b^{2})

*Example 6*

Factor xyz – x^{2}z

__Solution__

xyz – x^{2}z = xz (y – x)

*Example 7*

Factor 6a^{2}b + 4bc

__Solution__

6a^{2}b + 4bc = 2b (3a^{2} + 2c)

*Example 8*

Factor completely: x^{6 }– 64

__Solution__

x^{6 }– 64 = (x^{3})^{2} – 8^{2}

= (x^{3} + 8) (x^{3} – 8) = (x+2) (x^{2} − 2x + 4) (x − 2) (x^{2 }+ 2x + 4)

*Example 9*

Factor: x^{6 }– y^{6}.

__Solution__

x^{6 }– y^{6} = (x + y) (x^{2 }– xy + y^{2}) (x − y) (x^{2 }+ xy + y^{2})

### How to factor polynomials by grouping?

As the name suggests, factoring by grouping is simply the process of grouping terms with common factors before factoring.

*To factor a polynomial by grouping, here are the steps:*

- Check whether the terms of the polynomial have the Greatest Common Factor(GCF). If so, factor it out and remember to include it in your final answer.
- Break up the polynomial into sets of two.
- Factor out the GCF of each set.
- Finally determine if the remaining expressions can be factored any further.

*Example 10*

Factorize 2ax + ay + 2bx + by

__Solution__

2ax + ay + 2bx + by

= a (2x + y) + b (2x + y)

= (2x + y) (a + b)

*Example 11*

Factor ax^{2} – bx^{2} + ay^{2} – by^{2} + az^{2} – bz^{2}

__Solution__

ax^{2} – bx^{2} + ay^{2} – by^{2} + az^{2} – bz^{2}

= x^{2}(a – b) + y^{2}(a – b) + z^{2}(a – b)

= (a – b) (x^{2} + y^{2} + z^{2})

*Example 12*

Factor 6x^{2} + 3xy – 2ax – ay

__Solution__

6x^{2} + 3xy – 2ax – ay

= 3x (2x + y) – a (2x + y)

= (2x + y) (3x – a)

*Example 13*

x^{3} + 3x^{2} + x + 3

__Solution__

x^{3} + 3x^{2} + x + 3

= (x^{3} + 3x^{2}) + (x + 3)

= x^{2}(x + 3) + 1(x + 3)

= (x + 3) (x^{2} + 1)

*Example 14*

6x + 3xy + y + 2

__Solution__

6x + 3xy + y + 2

= (6x + 3xy) + (y + 2)

= 3x (2 + y) + 1(2 + y)

= 3x (y + 2) + 1(y + 2)

= (y + 2) (3x + 1)

= (3x + 1) (y + 2)

*Example 15*

ax^{2} – bx^{2} + ay^{2} – by^{2} + az^{2} – bz^{2}__Solution__

ax^{2} – bx^{2} + ay^{2} – by^{2} + az^{2} – bz^{2}

Factor out GCF in each group of the two terms

⟹ x^{2}(a – b) + y^{2}(a – b) + z^{2}(a – b)

= (a – b) (x^{2} + y^{2} + z^{2})

*Example 16*

Factor 6x^{2} + 3x + 20x + 10.

__Solution__

Factor out the GCF in each set of two terms.

⟹ 3x (2x + 1) + 10(2x + 1)

= (3x + 10) (2x + 1)

*Practice Questions*

Factor by grouping the following polynomials:

- 15ab
^{2}– 20a^{2}b - 9n – 12n
^{2} - 24x
^{3}– 36x^{2}y - 10x
^{3}– 15x^{2} - 36x
^{3}y – 60x^{2}y^{3}z - 9x
^{3}– 6x^{2}+ 12x - 18a
^{3}b^{3}– 27a^{2}b^{3}+ 36a^{3}b^{2} - 14x
^{3}+ 21x^{4}y – 28x^{2}y^{2} - 6ab – b
^{2}+ 12ac – 2bc - x
^{3}– 3x^{2}+ x – 3 - ab (x
^{2}+ y^{2}) – xy (a^{2}+ b^{2})

__Answers__

- 5ab (3b – 4a)
- 3n (3 – 4n)
- 12x
^{2}(2x – 3y) - 5x
^{2}(2x – 3) - 12x
^{2}y (3x – 5y^{2}z) - 3x (3x
^{2}– 2x + 4) - 9a
^{2}b^{2}(2ab – 3b + 4a) - 7x
^{2}(2x + 3xy – 4y^{2}) - (b + 2c) (6a – b)
- (x
^{2}+ 1) (x – 3) - (bx – ay) (ax – by)

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