how can i solve (x - 2)(2x - 5) = 0 by using factoring?

when 0 is the product of 2 numbers:
At least one of them must be 0
look; 6*0=0
0*15=0
0*0=0

In (x - 2)(2x - 5) = 0
(x - 2) = 0
x =2

or

(2x - 5) = 0
2x =5
x =5/2
x =2.5

(x - 2)(2x - 5) = 0
2x² - 5x - 4x + 10 = 0
2x² - 9x = - 10
x² - 9/2x = - 5
x² - 9/4x = - 5 + (- 9/4)²
x² - 9/4x = - 80/16 + 81/16
(x - 9/4)² = 1/16
x - 9/4 = 1/4

Factors of x:
= x - 9/4 - 1/4, = x - 10/4, = x - 5/2, = 2x - 5
= x - 9/4 + 1/4, = x - 8/4, = x - 2

Values of x:
2x - 5 = 0, 2x = 5, x = 5/2
x - 2 = 0, x = 2

Answer: x = 5/2, 2; (2x - 5)(x - 2) are the factors.

Proof (x = 5/2):
(5/2 - 2)(2[5/2] - 5) = 0
(5/2 - 4/2)(5 - 5) = 0
1/2(0) = 0
0 = 0

Proof (x = 2):
(2 - 2)(2[2] - 5) = 0
0(4 - 5) = 0
0(- 1) = 0
0 = 0

(x - 2)(2x - 5) = 0

x - 2 = 0
x = 2

2x - 5 = 0
2x = 5
x = 5/2 (2.5)

∴ x = 2 , 5/2 (2.5)

you don't factorise - it's already been done for you!

for (x-2)(2x-5) to equal 0, either (x-2)=0 or (2x-5)=0, thus either x=2 {(2-2)(2x-5)=0} or x=2.5 {(2.5-2)(2.5x2-5)=0). simple

x = 2 , x = 5 / 2

It is already factored...set each equal to 0 and solve for x (principle of zero products). x-2=0 and 2x-5=0 so...x=2 or x=5/2

equate each term to zero
you get
x-2= 0 and 2x-5=0
so x=2 and x=5/2 are the 2 solutions and factors