Question

Factorise
4a2-4b2+4a+1

Answer 1

Answer:

The required form is $4a^2-4b^2+4a+1=(2a+1-2b)(2a+1+2b)$

Step-by-step explanation:

Given : Expression $4a^2-4b^2+4a+1$

To find : Factorize the given expression?

Solution :

$4a^2-4b^2+4a+1$

Arranging the terms,

$=4a^2+4a+1-4b^2$

Making the square term,

$=[(2a)^2+2\cdot 2\cdot a+(1)^2]-4b^2$

The formula of square, $a^2+2ab+b^2=(a+b)^2$

$=(2a+1)^2-4b^2$

$=(2a+1)^2-(2b)^2$

Apply identity, $a^2-b^2=(a+b)(a-b)$

$=(2a+1-2b)(2a+1+2b)$

Therefore, The required form is $4a^2-4b^2+4a+1=(2a+1-2b)(2a+1+2b)$

Answer 2

Answer:

The factors obtained are $(2a+1+2b)$ and $(2a+1-2b)$.

Solution:

Given equation is $4a^2-4b^2+4a+1$

To factorise the given equation, we need to group the above variables into two groups:

$\begin{array} { c } { 4 a ^ { 2 } - 4 b ^ { 2 } + 4 a + 1 } \\\\ { = 4 a ^ { 2 } + 4 a + 1 - 4 b ^ { 2 } } \end{array}$

The above equation can be regrouped as:

$\begin{array} { c } { = ( 2 a ) ^ { 2 } + 2 \times 2 a \times 1 + 1 ^ { 2 } - ( 2 b ) ^ { 2 } } \\\\ { = ( 2 a + 1 ) ^ { 2 } - ( 2 b ) ^ { 2 } } \end{array}$

[Since, we know that the formula $( x + y ) ^ { 2 } = x ^ { 2 } + 2 x y + y ^ { 2 }$]

$= ( 2 a + 1 + 2 b ) ( 2 a + 1 - 2 b )$

[Since, we know that the formula $x^2+y^2=(x+y)(x-y)$]

Hence, $4a^2-4b^2+4a+1$ when factorised,  

The factors obtained are $(2a+1+2b)$ and $(2a+1-2b)$.