factorise 12x^3-3x?

Question

could anyone please show me the steps to solving this cubic equation by factorising it

Need Help!! Sos

Answer 1

This is not an equation as an equation requires
an " = " sign.
12 x ³ - 3 x
(3 x )(4 x ² - 1)
(3 x ) (2x - 1) (2x + 1)

IF this = 0 , values for x are then:-
x = 0 , x = 1 / 2 , x = - 1 / 2

Answer 2

For a quadratic equation such as the one above, you will have two factors in the form of (ax + b)... note, this is just the form for the factors. When we actually apply this form, we get: (Mx + P)(Nx + Q) When we multiply these two factors, the result is: (MNx^2) + (MQx) + (PNx) + (PQ) I wrote this out on two lines to better demonstrate how I multiplied both terms in the second factor first by the first term of the first factor, then by the second term of the first factor. You can actually multiply the terms in any order, but you must multiply both terms of each factor by both terms of the other. Placing these in a single line: (MNx^2) + (MQx) + (PNx) + (PQ) Next we can simplify this expression by adding the coefficients of the same variable with the same exponent: (MNx^2) + (MQ + PN)x + (PQ) From the original expression: MN = 3 PQ = 2 MQ + PN = -7 If we assume that the component factors of the coefficients are all whole numbers (we can not always assume this, but it is often the case for homework assignments), we get the following possibilities: MN = (1)(3) or (-1)(-3) PQ = (1)(2) or (-1)(-2) note that this means there can be FOUR possibilities for each pair, since the combinations can occur in any order. Solving for MQ + PN = -7, the easy possibilities for the two terms would either be -1 + -6 or -3 + -4 (again, these can be in any order). The second pair can be ruled out since neither MQ nor PN have any combinations that can produce 4. However, 1 + 6 is quite easy. MQ + PN = -7 MQ = (-3)(2) = 6 NP = (-1)(1) = 1 MQ = (3)(-2) = 6 NP = (1)(-1) = 1 Note that this time, order IS important! Thus: (Mx + P)(Nx + Q)= (-3x + 1)(-x + 2) = (1 - 3x)(2 - x) or (3x -1)(x -2) IMPORTANT: note that there are TWO possible answers! You can multiply these out to verify the answer.

Answer 3

Dear Ms. akta, p:


In using the processes of factorization of the equation
12x^3 - 3x = y.

letting y =0, we have the following:

12x^3 - 3x =0

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

3x = 0 or 2x + 1 =0 or 2x -1 =0


The solution set is { x: -1/2, 0 , 1/2 }



Sincerely,
[e-mail]
(a.k.a. calvaliear )

Answer 4

First take our common factors.
12x³-3x
3x(4x²-1) Then difference of squares.
3x(2x+1)(2x-1)

Answer 5

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

4x^2-1=0. x^2= 1/4. x= square root of 1/4.

Answer 6

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

Answer 7

12x^3 - 3x = 0

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

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

x = 0, -1/2, 1/2

Answer 8

12x^3-3x = 0
3x(4x^2 -1) = 0
3x(2^2x^2 - 1^2) = 0
3x(2x + 1)(2x - 1) = 0

Answer 9

=3x(4x^2-1)
x=0,1/2,(-1/2 not acceptable)