How do you factor #5y^2-12y+4=0#?

2 Answers
Aug 10, 2016

Answer:

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

Explanation:

Read all the clues which are given in the quadratic trinomial:

#color(turquoise)5y^2 color(orange)(-)color(magenta)(12)y color(red)(+)color(turquoise)4 = 0#

We need to find factors of #color(turquoise)(5 and 4)#

The factors of 5 and 4 must #color(red)"(ADD)"# together to give #color(magenta)(12)#

The signs in the brackets will be #color(red)"THE SAME"#
They will BOTH be #color(orange)"(NEGATIVE)"#"

Use factors of 5 and 4 and test the cross-products:

#color(white)(.....) 5 " "2 " " rArr 1xx2 = 2#
#color(white)(.....) 1 " "2" " rArr 5xx2 = 10" " 10+2 =color(magenta)(12)#

The top row gives the factors in one bracket.
The second gives the factors in the other bracket.

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

Solving would lead to #x = 2/5 or x = 2#

Aug 10, 2016

Answer:

(5x - 2)(x - 2)

Explanation:

Use the new AC Method (Socratic Search)
#f(y) = 5y^2 - 12y + 4 = 5(x + p)(x + q)#
Converted trinomial #f'(y)' = y^2 - 12 y + 20 =# (x + p')(x + q').
Method. Find p' and q' then divide them by a to get p and q.
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 20) --> (2, 10)(-2, -10). This last sum is -12 = b.
Then, p' = -2 and q' = -10
Back to f(y) --> #p = (p')/a = -2/5#. and #q = (q')/a = -10/5 = -2#.
Factor form of f(y):
#f(y) = 5(x - 2/5)(x - 2) = (5x - 2)(x - 2)#