Verified answer

3(x+y)^2-13(x+y)+4

Step1

substitute x+y by 'a'

given expression

=3a^2-13a+4

Step 2 Factorise

=3a^2-12a-a+4

=3a(a-4)-1(a-4)

=(a-4)(3a-1)

Step 3

Putting back the value of a

(x+y-4)(3x+3y-1) ans

1) let u = x+y

2) equation is now 3 u^2 - 13 u + 4

3) = (3u - 1) ( u - 4)

is this equation is equal to 0

then 4) 3 u = 1 and u = 4 are the possible answers

or u = 1/3 and u = 4

4) u = x + y so

x + y = 1/3 or x + y = 4

3(x+y)^2-13(x+y)+4

= [3(x+y) - 1][(x+y) - 4]

3(x+y)^2-13(x+y)+4

substitute x+y by 'a'

given expression

=3a^2-13a+4

Step 2 Factorise

=3a^2-12a-a+4

=3a(a-4)-1(a-4)

=(a-4)(3a-1)

Step 3

Putting back the value of a

(x+y-4)(3x+3y-1)

Let x+y = a

3a^2 - 13a + 4

(3a -1)(a - 4)

Subst

[3(x+y) - 1][(x+y) - 4]