try to solve it for factoring, then checking your answer using any other method. If you can't factor it but you can solve it using another method, how do you put the equation in its factored form.

24X²-14X-3 = 0

24X²-14X-3 = 0

24X²-14X-3 = 0

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Paoli, PA

Hi Harry,

This equation can be solved by factoring.

When i am doing these, i like to think of all the possible factors of the coefficient in front of x^2.....in this case 24.

Here is what they could be:

1 x 24

2 x 12

3 x 8

4 x 6

then, i do the same for the constant (3 in this case). Since 3 is a prime number, there is only one set of factors:

1 x 3

now the "hard" part is choosing which factors to use to get the middle coefficient (-14 in this case)

Since it is negative, we know that we will need a + and a - in the two factors:

(?x + ?)(?x - ?)

a little experimenting and we can get:

(6x + 1)(4x - 3) = 0

If we generalize this equation to be A * B = 0. So if we make A = 0 then B doesn't matter because 0 * anything = 0. Same idea for B. So what we do next is to make each factor a little mini equation like this:

(6x + 1) = 0

6x = -1

x = -1/6

(4x-3) = 0

4x = 3

x = 3/4

so the roots are:

x=3/4 or x = -1/6

Does that make sense?

Now, the harder part.....can every quadratic equation be factored? I believe the short answer is yes, but it may be very difficult if the coefficients are not easy numbers. However, there are different methods that you could use to figure it out (like completing the square and the quadratic formula ). These methods are not as simple, but will work regardless of how "easy" the factors are.

Woodland Hills, CA

aX^{2 }+ bx +c can be factored to , to binomial of ( X_{1} -a ) ( X_{2}-b) , if and and only if

X_{1 }+ X_{2 }_{= }-b/a

X _{1 * }X_{2 = }c/a_{ }a, b, c be all integers

Bolingbrook, IL

Every quadratic can technically be factored, but a lot of them can't be factored easily and not by the human brain without prior work. Like x^{2} + 5x -7 = 0. This can be factored like (x + )(x - ) = 0 but almost no one can just look at it and know what would go after the + and the minus. You would have to us the quadratic formula or complete the square and solve for x. Then the answers you get can go in the blanks, but by then there isn't a point, since the goal of factoring is to solve for x, and you already did with the quad.form or completing the square.

anyways to your question:

(6x + 1 )(4x - 3 ) = 0 so x = -1/6 and 3/4

**I did this by knowing the only factors of 3 are 3 and 1, so my only choices were

( +3)( - 1) or ( - 3)( +1). then I ran through the factors of 24, {1,2,3,4,6,8,12,24}. One of these bust be multiplied by the 3 and it's pair by the 1. Then they need to add or subtract to get -14.

3 x 4 or 3 x 6 seemed to be the closest to 14, but 3x4's counterpart is 1 x 6 which won't give you 14 anyway you work it. So 3 x6 has the counter part 1 x 4, which is 18 and 4 and you can get 14 by subtracting them or adding 18 and -4, but we want -14 so reverse the signs to -18 and 4. So I want a -3 and +1.

If you couldn't figure out the factoring arrangement, you could use the quad.form or complete the square. Both methods would give you x = -1/6 and 3/4.

Since when factoring we know the number that x is equal to need to plug into the parentheses to make 0, you can say

x + 1/6 x - 3/4 , if you multiply the first group by 6 and the second group by 4, that will get rid of the fractions and we will get, 6x +1 and 4x - 3, thus making (6x + 1 )(4x - 3 ) = 0

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