Algebraic proof?

Question

Show that:
(2a – 1)² - (2b – 1)² = 4(a – b)(a + b – 1)

Answer 1

Expand both terms.

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

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

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

The key step is to recognize the terms a^2 and -b^2 as being terms in the difference of two squares. From there, factoring the right pairs of terms reveals the final factored form.

Answer 2

(2a-1)^2 - (2b-1)^2
=4a^2 + 1 - 4a - 4b^2 -1 + 4b
=4a^2 - 4b^2 - 4a + 4b
=4(a^2 - b^2 - a + b)
=4{(a-b)(a+b)-(a-b)}
=4{(a-b)(a+b-1)} [by taking (a-b) common

Answer 3

First multiply everything out:

4a² - 4a + 1 - 4b² + 4b - 1 = 4(a² - a + b - b²)
4a² - 4a + 4b - 4b² = 4a² - 4a + 4b - 4b²

QED

Answer 4

x^2-y^2 = (x-y)(x+y) so

being
x=(2a-1)^2 and
y = (2b-1)^2

then

[(2a-1)+(2b-1)] [(2a-1)-(2b-1)]
(2a-1+2b-1)(2a-1-2b+1)
(2a+2b-2)(2a-2b)
2(a+b-1)(a-b)

Answer 5

easy
do it yourself then you wont have to ask questions on yahoo anymore to do your homework