I got 3(24x^2+36x-13) but can you make it 3(__x+__)(__x-__) fill in the blanks please..
2007-10-10 11:58:01 · 5 answers · asked by Allen C 3 in Science & Mathematics ➔ Mathematics
Hi,
If you mean can that be factored into real integers, then the answer is, no, it cannot. It's what we usually refer to as prime, at least in elementary studies of math.
A good way to tell if a quadratic equation can be factorized is to look at what is called the discriminant. That's √(b²-4ac), whare a, b, and c are taken from the equation
ax² +bx +c = 0. If the discriminant is not a perfect square; then the quadratic is prime.
In your case:
√(36²-4*24(-13))
=2544
Which is not a perfect square.
Hope this helps.
FE
2007-10-10 12:19:17 · answer #1 · answered by formeng 6 · 0⤊ 0⤋
No, I don't think so. You would need to find factors of 13x24 or 312 that subtract to give you 36. All the factors I come up with that even come close are 13 and 24, 26 and 12, 52 and 6, 39 and 8, and as you can see none of them have a difference of 36.
2007-10-10 12:07:37 · answer #2 · answered by Marley K 7 · 0⤊ 0⤋
3(24x² + 36x -13)
24 has the following factors: 1 and 24, 2 and 12, 3 and 8, 4 and 6
13 has only: 1 and 13
It sure looks like it would simplify more, but it doesn't, no matter what combination you use.
2007-10-10 12:10:59 · answer #3 · answered by lonedragonfli 2 · 0⤊ 0⤋
I find that it can be further factored only using radicals. By convention, that indicates that it is prime. The gimmick here is to equate the expression to zero and solve it with the quadratic formula; if the roots are R1 and R2, then the factorization is (x-R1)(x-R2). If the roots involve radicals or are complex, the expression is said to be prime.
2007-10-10 12:06:17 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Since b^2-4ac = (3)(4^2)(53), you can not fill in all rational numbers.
2007-10-10 12:05:40 · answer #5 · answered by sahsjing 7 · 0⤊ 0⤋