Factoring Quadratic equation?

Read the Steps and then follow the work.

ax^2 + bx + c = 0

1. Multiply a by c
2. Find numbers that multiply to ac and add/subtract to b (in your case we will subtract since the last sign in minus)
3. Rewrite b using the two numbers you found in step 2 (in your case one number needs to be positive and one negative, and since the first sign is positive the larger number needs to be positive)
4. Group the first two terms together, and group the last two terms together
5. Factor out the greatest common factor for each grouping (after you do this what is in the parenthesis shoudl be the same)
6. Make two new sets of parenthesis, the numbers outside each grouping go in one set, and the numbers inside each grouping go in the other set)
7. FOIL to get check your work.

3x^2 + 4x - 4 = 0
3(4) = 12 So we want number that mult to 12 and subtract to 4
6 (2) = 12 and 6-2 = 4

So now we rewrite replacing the middle term using 6 and 4
3x^2 - 2x + 6x - 4 = 0

Group the first two and the last two terms
(3x^2 - 2x) + (6x - 4) = 0

Now factor out the GCF of each group.
x(3x - 2) + 2(3x - 2) = 0

Now two new sets of parenthesis.
(x + 2)(3x - 2) = 0


Don't forget to FOIL to make sure your answer is correct
(x + 2)(3x - 2) = 0
3x^2 - 2x + 6x - 4 = 0
3x^2 + 4x - 4 = 0

If you want to solve for x you set each set of parenthsis = to 0
x + 2 = 0
x = -2

3x - 2 = 0
3x = 2
x = 2/3

I always use the formula. The solutions of any quadratic Ax² + Bx + C = 0 are
x = [-B + √(B² - 4AC)]/2A and x = [-B - √(B² - 4AC)]/2A
For your equation, A = 3, B = 4 and C = -4.
√(B² - 4AC) = √(16 + 48) = √64 = 8 and you get
x = (-4 + 8)/6 = 2/3 and x = (-4 - 8)/6 = -2

[The factors of 3x² + 4x - 4 are given by
3x² + 4x - 4 = 3(x - 2/3)(x + 2) = (3x - 2)(x + 2)]

3x^2 + 4x - 4 = 0
(3x - 2)(x + 2) = 0

3x - 2 = 0
3x = 2
x = 2/3

x + 2 = 0
x = -2

∴ x = 2/3 , -2

3x^2 +4x -4 =0. Remove the equation and times 'a' by 'c' where ax^+bx+c. The resulting equation will be x^2+4x-12. Then find 2 values where the product = 'c' (-12) and the sum = 'b'(4) You should find that the two values are 6 and -2. (6 times -2 = -12 and 6-2 = +4) Then substitute the values into the original equation where the 2 values replace 'b' and simply factorise: 3x^2 +6x -2x - 4 =0
3x(x+2) - 2(x+2) = 0
(3x-2)(x+2) =0

If you actually want to solve the equation rather than just factorising, make each individual bracket = 0:
3x-2 = 0 x+2 = 0
3x = 2 x = -2
x = 2/3

To check the two answers of x, substitute them back into the original equation.

3x^2+4x-4 =0
(3x-2)(x+2) = 0

3x-2 = 0 => x = 2/3;
x+2 = 0 => x = -2;

3x^2+4x-4 = 0

3 only has factors 3,1 while 4 only has factors 2, 2 and 4, 1
Since it's -4, then -2,2 are the most likely

(x+2)(3x-2)

x= -2 and/or x = 2/3

We first factor the left-hand side, getting

(3x - 2)(x + 2) = 0.

Thus, the product of the two quantities (3x - 2) and (x + 2) is zero. Now, if a product of two numbers is zero, it means that one or the other must be zero. In other words,

Either 3x - 2 = 0, giving x = 2/3,
or x + 2 = 0, giving x = -2.

Thus, there are two solutions: x = 2/3, and x = -2.

just use the Quadratic Formula....

(3x - 2)(x + 2) = 0
x = 2/3 , x = - 2

(3x-2)(x+2)=0
x = 2/3 and -2