Question

$Factorise (1+3y)^2 +(9y^2-1)$Factorise (1+3y)^2 +(9y^2-1)

Answer 1

Answer:

hey mate

(1+3y)²+(9y²-1)=((1)²+(3y)²+2×1×3y)+ (9y²-1)

=(1+9y²+6y) + (9y²-1)

=1+9y²+6y+9y²-1

= 18y²+6y

using identity ...

(a+b)= a²+b²+2ab

hope it will help you

Answer 2

$\sf {(1 + 3y)}^{2} + (9 {y}^{2} - 1) = 6y(3y + 1)$

Given :

$\sf {(1 + 3y)}^{2} + (9 {y}^{2} - 1)$

To find : To factorise

Solution :

Step 1 of 3 :

Write down the given expression

The given expression is

$\sf {(1 + 3y)}^{2} + (9 {y}^{2} - 1)$

Step 2 of 3 :

Simplify the given expression

$\sf {(1 + 3y)}^{2} + (9 {y}^{2} - 1)$

$\sf = {1}^{2} + 2.1 .3y + {( 3y)}^{2} + (9 {y}^{2} - 1)$

$\sf = 1+ 6y + 9{y}^{2} + 9 {y}^{2} - 1$

$\sf = 18 {y}^{2} + 6y$

Step 3 of 3 :

Factorise the expression

$\sf {(1 + 3y)}^{2} + (9 {y}^{2} - 1)$

$\sf = 18 {y}^{2} + 6y$

$\sf = 6y(3y + 1)$

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