How to factorise 12x^2-11x+2?

Factor by grouping is the method you must use.

(4 x - 1) (3 x - 2)

( 4x - 1 ) ( 3x - 2 )

Multiply 12*2 = 24

Find two factors of 24 that add up to -11

How about -8 and -3

(12x^2 + -8x )+ ( -3x + 2 )

Now factor out -4 from first bracket

12x^2-11x+2 = (4x - 1)(3x - 2) answer//

(4x-1)(3x-2)

When things get this large, I usually don't try to factor, but if I need to factor, I would solve for the roots and then turn the roots into factors:

12x² - 11x + 2

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -(-11) ± √((-11)² - 4(12)(2))] / (2 * 12)

x = [ 11 ± √(121 - 96)] / 24

x = [ 11 ± √(25)] / 24

Note that if what's under the radical isn't a positive rational number, this can't be factored over rational numbers.

x = (11 ± 5) / 24

x = 6/24 and 16/24

x = 1/4 and 2/3

So we get:

(x - 1/4) (x - 2/3)

Recall this would be in an equation equal to 0 if we were factoring to solve for zeroes, so let's do that:

(x - 1/4) (x - 2/3) = 0

Now the only problem is in factoring, we wouldn't have fractions. So we can multiply each factor by its denominator (and multiply the right side by the same values), so multiply by 4 and 3:

4(x - 1/4) * 3(x - 2/3) = 0 * 4 * 3

(4x - 1)(3x - 2) = 0

So the factored form of your expression is:

(4x - 1)(3x - 2)

make it equal zero

12x² - 11x +2 = 0

solve this equation

ax² + bx + c = 0

b rule

{11 ± √ [ 11² - (4)(12)(2)] } / 24 =

(11 ± 5) / 24

16/24 =2/3 and 6/24 = 1/4

now (a) in front and changing the signs

12(x - 2/3 ) ( x - 1/4 ) =

(3x-2)(4x-1)

12x²-11x+2

12x²-(11+d)x+dx+2

Here we want 12:-(11+d) = d:2 in order to get a common factor, so we have d²+11d+24=0. This should be more straightforward to solve by your usual methods for d (-3 or -8). Substitue one of these back into our equation.

12x²-(11-3)x-3x+2

12x²-8x-3x+2

4x(3x-2)-(3x-2)

(4x-1)(3x-2)

12x²-11x+2=(4 x - 1) (3 x - 2)

x²-(11/12)x+(2/12)= etc,.