Question

Factorise:
$(4x ^{2} - 4x + 1) - 9y ^{2}$
Using identity:
$x ^{2} - y ^{2} = (x + y)(x - y)$
​

Answer 1

Answer:

We know that

(x

2

+y

2

+z

2

+2xy+2yz+zx)=(x+y+z)

2

(i) 4x

2

+9y

2

+16z

2

+12xy−24yz−16xz

=(2x)

2

+(3y)

2

+(−4z)

2

+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)

=(2x+3y−4z)

2

=(2x+3y−4z)(2x+3y−4z)

(ii) 2x

2

+y

2

+8z

2

−2

2

xy+4

2

yz−8zx

=(−

2

x)+(y)

2

+(

2

z)+2(−

2

x)(y)+2(y)(2

2

z)+2(2

2

z)(y)

=(−

2

x+y+2

2

z)

2

=(−

2

x+y+2

2

z)(−

2

x+y+2

2

z)

Answer 2

Step by step explanation:
Given: (4x2 - 4x + 1) - 9y2

Using x2 - y2 = (x + y)(x - y) identity

We rewrite the equation (4x2 - 4x + 1) as (-1 + 2x)^2 and 9y2 as (3y)^2

So (-1 + 2x)^2 - (3y)^2

As square is common we take the square out and combine the equation

= (-1 + 2x - 3y)^2

= (-1 + 2x - 3y) (-1 + 2x + 3y)

Reordering the terms

= (2x - 3y - 1) (2x + 3y - 1)