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

Please give steps as well as answer. I am doing this for revision. I can't find anything that would help. The method I remember being taught doesn't seem to work at all, although it worked with expressions with a larger number than x.

'expressions with a larger number than x' example being '12x^2-11x+30' or something.

### 11 Answers

- Jeffrey KLv 72 years ago
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

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- llafferLv 72 years ago
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)

- khalilLv 72 years ago
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)

- Φ² = Φ+1Lv 72 years ago
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)