Factor completely?

Question

I never learned how to do a problem like this:

12x^2 + 57xy - 21y^2

its written 57 on my paper

Answer 1

The factors are of the form (ax + by) * (cx - dy)

so ac = 12 so 12, 1 or 4. 3 or 3, 4 or 1, 12

bd = 21 again - four combinations


cb - da = 57

I can't see that it works. Are you sure you wrote it correctly ? A 59 instead of 57 would work. 3 * 21 - 4 * 1 = 63 - 4 = 59

Answer 2

I worked this problem without success. It got so bad that I canceled my efforts, because the answer I got was absolutely baffling.
I then checked out the answerers who responded to you. What follows are my views on their replies.
DZ says the answer is (4x+21y)(3x-y). That answer is incorrect, although I suspect he is going to be proven right eventually. More about that later.
If you expand (4x+21y)(3x-y), you will get
12x^2+63xy-4xy-21y^2, =12x^2+59xy-21y^2. This is not 12x^2 +57xy-21y^2
Beardo asks if you are sure the middle term isn't 59, because if it was, factoring would work. He's right. In fact, if it is 59 instead of 57, DZ has the correct answer.
Profmark went the quadratic formula route and got a messy answer. So did I, but I now think you cannot use the quadratic formula on this problem. It contains 2 variables, and the formula is for only 1 variable.
He too suggests the middle term might be wrong, and that 59 would work.
Reginald takes out the common factor 3, to get
3(4x^2+19xy-7y^2), and leaves it at that. He is the
closest to the only answer that makes sense.

I think you need 59 for the middle term, or that you have a really nasty textbook!

Answer 3

12x² + 57xy - 21y² = 0

The middle term is + 57xy

Find the sum of the middle term

Multiply the first term 12 times the last term 21 equals 252 and factor

factors of 252

1 x 252
2 x 126
3 x 84
4 x 63. . .←. .use these factors
6 x 42
7 x 36
9 x 28
12 x 21
14 x 18

+ 63 and - 4 satisfy the sum of the middle term

insert + 63xy and - 4xy into the equation

12x² + 57xy - 21y² = 0

12x² - 4xy + 63xy - 21y² = 0

Group factor

(12x² - 4xy) + (63xy - 21y²) = 0

4x(3x - y) + 21y(3x - y) = 0

(4x + 21y)(3x - y) = 0

- - - - - - - -s-

Answer 4

To factor completely, first remove any common monomial
factor. All the terms are divisible by 3 so we can say:
12x² + 57xy - 21y² = 3(4x² + 19xy - 7y²)

Answer 5

Check that the middle term is not supposed to be 59xy. If it is then there is an integer solution:
(4 x + 21 y) (3 x - y)
since 21*3-4*1 = 63-4 = 59 NOT 57 !

As given, the solution is rather nasty. Using the quadratic formula you get:
***CORRECTED***
12 (x + ((19+sqrt(473))/8) y) (x + ((19-sqrt(473))/8) y)
***CORRECTED***

For details on how to do this factorization and to see the multiplication check see http://www.geocities.com/timbermusic2002/mark/factor-problem.pdf

Answer 6

(4x+21y)(3x-y)

Answer 7

yup he's correct.