Factorization - 1 | Lesson Note

Intro

What is Factorization?

Let us consider a number. So, 6 is a result of multiplication of 2 and 3.

i.e. $2 \times 3 = 6$

The process of getting 2 and 3 from 6 is called factorization.

Factorization is a reverse process of multiplication.

A

Factorization of Monomials (Having one term)

i) $2x = 2 \cdot x$
ii) $3x^2 = 3 \cdot x \cdot x$
iii) $12xy = 2 \cdot 2 \cdot 3 \cdot x \cdot y$
iv) $-14xy^2 = -2 \cdot 7 \cdot x \cdot y \cdot y$
v) $25 = 5 \cdot 5$
vi) $(a+b)^2 = (a+b)(a+b)$

B

Factorization of Binomials (Having two terms)

$\therefore$ 2 and (x+2) are factors of $2x+4$.

ii) $x^2 + 2x$

= x $\cdot x$ + $2 \cdot$ x

= $x(x + 2)$

$\therefore x$ and $(x+2)$ are factors of $x^2 + 2x$.

iii) $-6 + 2x$

[shuffle the terms, keep negative inside]

= $2x - 6$

= 2 $\cdot x - $ 2 $\cdot 3$

= $2(x - 3)$

iv) $-2x^2 + 4x$

= $4x - 2x^2$

= 2 $\cdot 2 \cdot$ x $-$ 2 $\cdot x \cdot$ x

= $2x(2 - x)$

v) $10n + 5$

= $2 \cdot$ 5 $\cdot n +$ 5

= $5(2n + 1)$

Note:
• If nothing is common, 1 is common.
• If nothing is remaining, 1 is remaining.

C & D

Factorization of Trinomials & 4-terms

C) Factorization of Trinomials (Having 3 terms)

i) $ax^3 + bx^2 + cx$

= $a \cdot$ x $\cdot x \cdot x + b \cdot$ x $\cdot x + c \cdot$ x

= $x(ax^2 + bx + c)$

ii) $2x^2 - 6x + 4$

= 2 $\cdot x \cdot x -$ 2 $\cdot 3 \cdot x + $ 2 $\cdot 2$

= $2(x^2 - 3x + 2)$

☞ In trinomial common value must be present on all three terms.

D) Factorization of 4-terms (Rearranging and grouping)

i) $ax + ay + bx + by$

Rearranging in 'x' and 'y'

= $ax + bx + ay + by$

= $a \cdot$ x $+ b \cdot$ x $+ a \cdot$ y $+ b \cdot$ y

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

common value $\downarrow$

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

Important: If there are 4 terms in an algebraic expression then we have to take a group of 2 terms each to factorize.

ii) $x^2 + 3n + 2n + 6$

= x $\cdot x + 3 \cdot$ x $+ $ 2 $\cdot n + $ 2 $\cdot 3$

= $x(x+3) + 2(x+3)$

= $(x+3)(x+2)$

E

Difference of Squares

Activity:

Fig-(I)

Fig-(II)

Fig-(III)

Fig-(IV)

Observation (Figure explainer):

A piece (square) of length 'a' is taken. An attempt to cut a square piece of length 'b' is shown. [Fig-I]
A square piece of length 'b' is taken out from square piece of length 'a'. [Fig-II]
From the remaining part again a piece is attempted to cut out. This time length a-b and width b. [Fig-III]
A piece of length a-b and width 'b' is taken out and adjusted along the length a-b of remaining piece. [Fig-IV]

Conclusion:

Fig-I
length (l) = a
breadth (b) = a
Area $(A_1) = a \times a = a^2$

Fig-II
A piece of $b^2$ is taken out,
so,
Area $(A_2) = a^2 - b^2$

Fig-IV
length (l) = a + b
breadth (b) = a - b
Area $(A_4) = l \times b$
$= (a+b)(a-b)$

Area of fig (II) and Area of fig (IV) must be same.

So, $A_2 = A_4$

$\therefore a^2 - b^2 = (a+b)(a-b)$

Simple examples:

a) $x^2 - 4$

• First express as difference of square: $(x)^2 - (2)^2$

• Whatever is written inside brackets are 'a' and 'b'.
So, $a = x$, $b = 2$, apply $(a+b)(a-b)$

= $(x+2)(x-2)$

b) $25 - z^2$

• First express as difference of square: $(5)^2 - (z)^2$

• Whatever is written inside brackets are 'a' and 'b'.
So, $a = 5$, $b = z$, apply $(a+b)(a-b)$

= $(5+z)(5-z)$

c) $x^2 - \frac{1}{x^2}$

• First express as difference of square: $(x)^2 - (\frac{1}{x})^2$

• Whatever is written inside brackets are 'a' and 'b'.
So, $a = x$, $b = \frac{1}{x}$, apply $(a+b)(a-b)$

= $(x+\frac{1}{x})(x-\frac{1}{x})$

d) $(x-y)^2 - 16$

• First express as difference of square: $(x-y)^2 - (4)^2$

• Whatever is written inside brackets are 'a' and 'b'.
So, $a = x-y$, $b = 4$, apply $(a+b)(a-b)$

= $(x-y+4)(x-y-4)$

Complex examples:

a) $x^3y - xy^3$

• First look for common, do like normal binomial.

= x $\cdot x \cdot x \cdot $ y $-$ x $\cdot y \cdot y \cdot $ y

= $xy(x^2 - y^2)$

• Second express as difference of square.

= $xy\{(x)^2 - (y)^2\}$

= $xy(x+y)(x-y)$

b) $2x^2 - \frac{18}{y^2}$

• First look for common, do like normal binomial.

= 2 $\cdot x^2 - $ 2 $\cdot \frac{9}{y^2}$

= $2(x^2 - \frac{9}{y^2})$

• Express as difference of square.

= $2\{(x)^2 - (\frac{3}{y})^2\}$

• Identify 'a' and 'b', apply $(a+b)(a-b)$.
i.e. $a = x$, $b = \frac{3}{y}$

= $2(x+\frac{3}{y})(x-\frac{3}{y})$