Subject: Compulsory Maths

Factorization is the process of finding the factors. Factoring is the decomposition of an object, into a product of other objects, or factors, which when multiplied together give the original. This note contains information about factorization.

When two or more algebraic expressions are multiplied, the result is called product and each expression is called the factor of the product.

The process of finding out factors of an algebraic expression is known as factorisation.

For example:

If we factorise (bc + cd), you get c ( b + d ).

Let's multiply ( a + b ) and ( a - b )

( a + b ) ( a - b )

= a² - ab + ab -b²

= a² - b² ( This expression is called a difference of two squares )

Therefore, the factors of a² - b² are ( a + b ) and ( a - b)

**Examples:**

- x
^{2 }- 49

Solution:

x^{2 }- 49, this expression is the difference of two squares.

= x^{2}- 7^{2}, which is in the form of a^{2}- b^{2}= (x+7) (x-7) - 4y
^{2 }- 36y^{6}Solution:

In 4y^{2 }- 36y^{6}, there is a common factor of 4y^{2}that can be factored out first in this problem, to make the problem easier.

= 4y^{2 }- 36y^{6}= 4y^{2}(1 - 9y^{4})

= 4y^{2}{(1)^{2}- (3y^{2})^{2}}

= 4y^{2}(1+3y^{2})(1-3y^{2})

Let's multiply (a+b) and (a+b)

(a+b) (a+b)

= a^{2}+ ab + ab + b^{2}

= a^{2} + 2ab + b^{2}

Thus, a^{2} + 2ab + b^{2} = (a + b)^{2} and (a + b)^{2} is the factorisation form of a^{2 }+ 2ab + b^{2}

Similarly, a^{2 }- 2ab + b^{2} = (a -b)^{2} and (a - b)^{2} is the factorisation form of a^{2 }- 2ab +b^{2 }= (a -b)^{2 }and (a - b)^{2} is the factorisation form of a^{2} - 2ab + b^{2}

If we consider (a+ b) as one of the side of the square then the product of the expression will form two squares namely a^{2} and b^{2} and two congruent rectangles with each having an area of ab.

a^{2} |
ab |

ab | b^{2} |

Area of the entire square = (a + b)^{2}

Area of two squares and two rectangles

= a^{2} + ab +ab + b^{2}

= a^{2} + 2ab +b^{2}

Thus, a^{2 }+ 2ab + b^{2} = (a+b)^{2}

- Factorization is the process of finding the factors.
- Factoring is the decomposition of an object, into a product of other objects, or factors, which when multiplied together give the original.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Factorise:

y^{2 }- 9

Solution:

y^{2}-9, this expression is the difference of two squares.

=y^{2}-3^{2}, which is in the form of a^{2}-b^{2}

=(y+3)(y-3)

Factorise:

4y^{2 }- 36^{6}

Solution:

In 4y^{2}-36^{6}, there is a common factor of 4y^{2} that can be factored out first in this problem, to make the problem easier.

4y^{2}- 36y^{6}

= 4y^{2}(1 - 9y^{4})

= 4y^{2}{(1)^{2}- (3y^{2})^{2}}

= 4y^{2}(1 + 3y^{2})(1 - 3y^{2})

Factorise: 6x+3

Solution:

Given expression =6x+3

= 2.3.x + 3

= 3(2x+1) [3 is common in both]

Factorise: x^{2}+4x

Solution:

Given expression =x^{2}+4x

=x. x+4. x

=x(x+4) [x is common in both]

Factorise: 12a+3b

Solution:

Given expression =12a+3 b

=4.3. a + 3.b

=3(4a+b) [ 3 is common in both]

Factorise: x+x^{3}

Solution:

Here,

Given expression =x+x^{3}

=x+x . x.x

=x(1+x^{2}) [x is common in both]

Factorise: 12x^{2}+xy+xz

Solution:

Here,

Given expression =12x^{2}+xy+xz

=2.2.3.x.x+x.y+x.z

=x(12x + y + z) [ x is common in all]

Factorise: 14xy+7y

Solution:

Here,

Given expression =14xy+7y

=2.7.x.y+7.y

=7y(2x+1) [ 7y is common in both]

Factor the following trinomials.

9a^{2}+24ab+16b^{2}

Solution:

9a^{2}+24ab+16b^{2}

Since 24 =2×3×4 and 9=3^{2},16=4^{2},

9a^{2}+24ab+16b^{2}=(3a)^{2}+2.3a.4b+(4b)^{2}

=(3a)^{2}+2.3a.4b+(4b)^{2}

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

Factor the trinomials.

25a^{2}-80a+64

Solution:

25a^{2}-80a+64

Since 80=2×5×8, 25=5^{2} and 64=8^{2},

25a^{2}-80a+64=(5a)^{2}-2.5a.8+(8)^{2}

= (5a-8)^{2}

Factorise:

x^{2}+5x+6

Solution:

Comparing x^{2}+5x+6 with x^{2}+(a+b) x+ab we get,

a+b =5 and ab=6

So, 3+2=5 and 3×2=6

Hence, x^{2}+5x+6=x^{2}+(3+2)x+6

=(x+3)(x+2)

ab=6 | a+b=5 |

1×6=6 | 1+6=7 |

2×3=6 | 2+3=5 |

Factorise:

a^{2}+7a-18

Solution:

For a^{2}+7a-18, two numbers whose difference is 7 and product-18 are 9 and -2

Hence, a^{2}+7a-18=a^{2}+(9-2)a-18

= (a+9)(a-2)

Factorise:

b^{2}-7b-18

Solution:

For b^{2}-7b-18, two numbers whose sum is -7 and product -18 are -9 and 2

Hence, b^{2}-7b-18

=b^{2}(-9+2)b-18

= (b-9)(b+2)

Factor

64x^{3}+125

Solution:

64x^{3}+125=(4x)^{3}+5^{3}

=(4x+5)[(4x)^{2}-4x.5+5^{2}]

=(4x+5)(16x^{2}-20x+25)

Factor:

3a^{3}-81

Solution:

3a^{3}-81=3(a^{3}-27)

=3(a^{3}-3^{3})

=3(a-3)(a^{2}+3a+9)

Factorise:

x^{2}-7x=12

Solution:

Comparingx^{2}-7x=12 with x^{2}-(a+b) x+ab we get,

a+b=7 and ab =12

So, 4+3=7 and 4×3=12

Hence, x^{2}-7x+12=x^{2}-(4+3) x+12

=x^{2}+(-4-3)x + 12

= (x-4)(x-3)

ab=12 | a+b=7 |

12×1=12 | 12+1=13 |

6×2=12 | 6+2=8 |

4×3=12 | 4+3=7 |

Factorise the following by taking common:

x^{2}+3x+xy+3y

Solution:

Here givenx^{2}+3x+xy+3y

=x(x+3)+y(x+3)

=(x+3)(x+y) Ans.

Using a^{2}-b^{2} formula, factorise the following:

x^{2}-4

Solution:

Here,

x^{2}- 4

=(x)^{2}-(2)^{2}

=(x-2)(x+2) [\(\therefore\)a^{2}-b^{2}=(a+b)(a-b)]

Using a^{2}-b^{2} formula, factorise the following:

9x^{2}-y^{2}

Solution:

Here,

Given = 9x^{2}-y^{2}

=(3x)^{2}-(y)^{2}

=(3x+y)(3x-y)

Using a^{2}-b^{2} formula, factorise the following:

121-25y^{2}

Solution:

Given = 121-25y^{2}

= (11)^{2}-(5y)^{2}

= (11+5y)(11-5y)

Factorize the algebraic expression of the form x^{2} + px + q:

x^{2} - 7x + 12

Solution:

The given expression is x^{2} - 7x + 12

Find two numbers whose sum = -7 and product = 12

Clearly, such numbers are (-4) and (-3).

Now, x^{2} - 7x + 12

= x^{2} - 4x - 3x + 12

= x(x - 4) -3 (x - 4)

= (x - 4)(x - 3)

Factorize the algebraic expression of the form ax^{2} + bx + c:

15x^{2} - 26x + 8

Solution:

The given expression is 15x^{2} - 26x + 8.

Find two numbers whose sum = -26 and product = (15 × 8) = 120.

Clearly, such numbers are -20 and -6.

Here, 15x^{2} - 26x + 8

= 15x^{2} - 20x - 6x + 8

= 5x(3x - 4) - 2(3x - 4)

= (3x - 4)(5x - 2)

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