11 Answers

SOLUTION: (3x 5)(2x 1) = 0
EXPLANATION:
First notice the highest term coefficient. it is 6.
That means it can be either
(2x+___)(3x+___) //* or could be negative sign
or
(x+____)(6x+___)
Those would make highest term 6x^2.
Now focus on constant 5.
It can be
(2x__1)(3x___5)
Where ___ indicates you have to do trial and error for sign.
To get positive 5, it can be either both signs are positive or both are negative. But now look at the middle terms. It is 13
So in order to get 13, you need both negative signs because the middle term in binomial is where outside product and inside product in nested form gets added in FOIL method.
So it should be (3x 5 )(2x1)
The more you practice, you will be able to do it in your head. Otherwise you need to do many trials and errors.
Now in case you have to solve for x, it should be obvious that 3x5 = 0 and 2x 1 = 0
x = 5/3, 1/2

6x^2  13x +5 = 0
x^2  13x + 30 = 0
(x  10) (x 3) = 0
(x  10/6) (x 3/6) = 0
(x  5/3) ( x  1/2) = 0
(3x  5) ( 2x1) = 0

6x^213x+5=0
(2x 1 )(3x 5 )=0
2x1=0 or x5=0
2x=1 or x=5
x=1/2 or x=5
the solution set ={1/2 , 5}

=> 6x^213x+5 = 0
=> 6x^23x10x+5 = 0
=> 3x(2x1)5(2x1) = 0
=> (3x5)(2x1) = 0
Either, x = 5/3, or x = 1/2

6x^2  13x + 5 = 0
(3x5)(2x1) = 0

6x²  13x + 5 = 0
6x²  3x  10x + 5 = 0
3x(2x  1)  5(2x  1) = 0
(2x  1)(3x  5) = 0
x = 1/2, 5/3

6x^213x+5
if the second operation is +,
then both operations in solution
are the same
if first operation determines the operations in the solution ()
6x^213x+5
(  ) (  )
6x^213x+5
possible solutions:
(6x ) (1x )
(1x ) (6x )
(3x ) (2x )
(2x ) (3x )
^ and ^ must be 1 & 5 because
they are factors of 5 in 6x^213x+5
Possible solutions:
(6x1)(x5)
(x1)(6x5)
(3x1)(2x5)
(2x1)(3x5)
Use FOIL (first, outside, inside, last) . "first" & "last" is done
"outside" & "inside" added up must be 13x
explanation of FOIL: (a+b)(a+b)=aa+ab+ab+bb=a^2+2ab+b^2
aa(first) ab(outside) ab(inside) bb (last)
(6x1)(x5)
30xx nope
(x1)(6x5)
6x5x 11x nope
(3x1)(2x5)
2x15x nope
(2x1)(3x5)
3x10x = 13x correct
The answer is (2x1)(3x5)
(2x1)(3x5)=0
Solve for x
2x1=0 or 3x5=0
+, , x, or / by both sides of the equation
2x=1 or 3x=5
2/2x=1/2 or 3/3x=5/3
1x=1/2 or 1x=5/3
x=1/2 or x=5/3
x={1/2, 5/3}

(3x5)(2x1)=0
x = 5/3, 1/2

use the quadratic formula (B +/ sqrt(b^2  4AC))/2a
where
A= 6
B= 13
C= 5
(13 +/ sqrt(13^2  4(6)(5)))/2
so you get two answers
x = .5 and x = 1.6667

(3x+1)(2x5)=0
x=1/3
x=5/2