Algebra: Expansion and Factorization

Welcome Back to My Blog!

Today, we're diving into the fascinating world of algebra. Get ready as we embark on this exciting adventure together!

 The Importance of Algebra 

Algebra is essential in various fields of mathematics. It involves manipulating equations and solving problems with unknown variables. In this chapter, we’ll focus on expanding expressions involving brackets and the reverse process called factorization. In expansion,we remove brackets but in factorization,we introduce them.

 The Distributive Law 

Consider the expression a(b+c). Here, a is the coefficient of the expression in the brackets,(b+c) . The distributive law says that to simplify a(b+c), we must multiply  the coefficient with each term inside the brackets and add the results:

a(b+c) = ab + ac

Example 1:

1. 2(5x-1)=2(5x)+(2×(-1))

      =10x-2

2. 2x(3-x)=(2x × 3)+(2x ×(-x))

   =6x-2x²

3. -5x(x-3) =(-5x × x)+(-5x ×(-3))

      = -5x² + 15x

Example 2:Let’s expand and simplify the following expressions.

1. 2(5x-1) + 3(2-x)

Expands to: 10x-2+6-3x

Simplifies to: 7x+4

2. x(x-2)-3x(5-x)

Expands to: x²-2x-15x+3x²

Simplifies to: 4x² -17x

 The Product of Two Binomials 

Consider the expression, (w+x)(y+z). To expand it, we multiply each term in the first bracket by each term in the second bracket:

(w+x)(y+z) = w(y+z)+x(y+z)

=wy + wz + xy + xz

Example 3:

1. (2x-1)(x+3)

Expands to: 2x(x+3)-1(x+3)

=2x² +6x-x-3

=》2x² +5x-3

2. (x+1)(x+2)

Expands to: x(x+2)+1(x+2)

=x² +2x+x+2

=x² +3x+2

 Difference of Two Squares 

Consider the expression x²-y², x²  and y² are termed as perfect squares and so the expression x²-y² is called difference of two squares.It can be factored as (x-y)(x+y).Therefore,x²-y² is the expanded form of  (x-y)(x+y)

Proof:

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

=x²-xy+xy-y²

=x²-y²

 Expanding Perfect Squares 

Expressions like (x+y)² and (x-y)² are termed as perfect squares.

 Rules for expanding perfect squares 

Step 1;Square the first term.

2. Add twice of the product of the first and last terms.

3. Add the square of the last term 

Example 5:

1. (x+y)²=x² +(2× x ×y)+y²

     =x² +2xy +y²

2. (4-3x)=4² +(2×4×(-3x))+(-3x)²

     =16-24x+ 9x²

 Factorization with Common Factors 

Factorization is the process of writing an expression as a product of its factors. This is the reverse process of expansion since in factorization, we introduce brackets but in expansion,we remove them.

Example:

1. 3a+6

3 is a common factor of both 3a and 6,so we factorize 3 out.

=》3(a+2)

2. 9x² +12x.We factorize the common factor,3x out.

    =》3x(3x+4)

 Special Cases: Difference of Two Squares 

As mentioned earlier,the introduction of brackets is factorization.

Example:Factorize the following completely.

1. 9-x²=3²-x²

     =(3+x)(3-x)

2. 4x² -25 =2²x² -5²

     =(2x)²- 5²

     =(2x+5)(2x-5)

 Factorizing Quadratic Expressions 

Quadratic expressions take the form ax²+bx+c . To factor ax²+bx+c , we find two numbers that multiply to  give us ac and add to b.

Consider the expression x²+3x+2. It can be written in the form (x+a)(x+b),

where a and b are numerous that add up to give 3 and multiply to give us 2×1 which is equal to 2.In this case,these numbers are 2 and 1.

Therefore,x²+3x+2 factors to (x+1)(x+2)

Example;Factorize the following quadratic expressions.

i.x²+11x+24

a×c=1×24=24

Factors of 24={1,2,3,4,6,8,12,24}.

Two factors which when added gives us 11 and when multiplied gives us 24 are 8 and 3.

=》x²+11x+24

=(x+8)(x+3)

ii.x²-7x+12

Since the sum of the roots is negative and the product of the roots is positive,we're looking for two negative numbers since (-×-)=+ and (- + -)= -.

Negative factors of 12={-1,-2,-3,-4,-6,-12}.We notice that -3 and -4 sum up to give us -7 And their product gives us +12.

=》x²-7x+12

=(x-4)(x-3)

iii.3x²+ 6x-72

In this case,we notice that 3 is a common factor to all 3 terms in the expression. Therefore,we factorize 3 out and work with whatever will be left in the bracket.

=》3x²+ 6x-72

=3(x²+2x -24)

Since the sum of the roots is positive and the product of the roots is negative,we're looking for two numbers,a positive and a negative number where the positive number is greater than the negative number.

ac=-24×1= -24.Let's determine the positive and positive and negative factors of 24.

Positive factors of 24={1,2,3,4,6,8,12,24}

Negative factors of 24={-1,-2,-3,-4,-6,-8,-12,-24}.

The two numbers that can satisfy our condition are 6 and -4.

=》3x²+ 6x-72

=3(x-4)(x+6)

 Assignment 

1.Expand and simplify 

i.(3x-y)^2

ii.-2a(b-a)

iii.-(5-x)^2

iv.2(x+3)(x+2) - 3(x+2)(x-1)

v.(x+4)(x^2 -x+2)

2.Factorize the following completely.

i.5x-5 +xy-y

ii.2x^2-98

iii.6-5x+x^2

iv. xy+2x+2y+4

v.(x-y)a +(x-y)

vi.(x-2)y -(x-2)z.

Feel free to add your thoughts or ask questions in the comments below! Happy learning!