Factor 6x^2 – 13x + 5 = 0?

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 )(2x-1)

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 3x-5 = 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) ( 2x-1) = 0

6x^2-13x+5=0

(2x -1 )(3x -5 )=0

2x-1=0 or x-5=0

2x=1 or x=5

x=1/2 or x=5

the solution set ={1/2 , 5}

=> 6x^2-13x+5 = 0

=> 6x^2-3x-10x+5 = 0

=> 3x(2x-1)-5(2x-1) = 0

=> (3x-5)(2x-1) = 0

Either, x = 5/3, or x = 1/2

6x^2 – 13x + 5 = 0

(3x-5)(2x-1) = 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^2-13x+5

if the second operation is +,

then both operations in solution

are the same

if first operation determines the operations in the solution (-)

6x^2-13x+5

( – ) ( – )

6x^2-13x+5

possible solutions:

(6x- ) (1x- )

(1x- ) (6x- )

(3x- ) (2x- )

(2x- ) (3x- )

^ and ^ must be 1 & 5 because

they are factors of 5 in 6x^2-13x+5

Possible solutions:

(6x-1)(x-5)

(x-1)(6x-5)

(3x-1)(2x-5)

(2x-1)(3x-5)

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)

(6x-1)(x-5)

-30x-x nope

(x-1)(6x-5)

-6x-5x -11x nope

(3x-1)(2x-5)

-2x-15x nope

(2x-1)(3x-5)

-3x-10x = -13x correct

The answer is (2x-1)(3x-5)

(2x-1)(3x-5)=0

Solve for x

2x-1=0 or 3x-5=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}

(3x-5)(2x-1)=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)(2x-5)=0

x=-1/3

x=5/2