i have a doubt in maths_quadratic equations_factorisation?

Question

plz tell me how to easily factorise an equation involving big numbers for ex. this one
3n^2-n-1430
plz teach me an easy method to do this
1'm s'posed to complete the full sum within 5 minutes
tanx a dozen times
tom

Answer 1

Multiply 3 and -1430
3(-1430) = -4290
You're looking for factors of -4290 that add to -1 (which is the coefficient of the n term)
-66(65) = -4290 and -66 + 65 = -1
Replace -n with -66n + 65n
3n^2 - 66n + 65n - 1430
3n(n-22) + 65(n-22)
= (n-22)(3n+65)

Answer 2

Mathgirl's answer looks pretty good, although I didn't check it. I screwed up the math and quit. (Oh, I see, I forgot that the 3 came from the "x" coefficient).

I can't remember if she said WHY you multiply by three, but it is to make the coefficient of the first terms the same, so you have less to be concerned about during your factorization.

I would add these comments:
Look for your coefficients systematically.

The prime factorization of 1430*3=2*3*5*11*13.
so there are a LOT of combinations.

Notice that the closer the two factors in absolute value, the smaller the difference is, Since our middle term is small, we want factors close to the square-root of the number.

I didn't have a calculator, so I did trial and error, my thinking process is detailed below.

In general, in factoring a trinomial, I like to arrange my factors in a table, and leave room to fill in as I need to.
Like this:

First Second Difference
65, 66,1
55,78,-
30,143,-
3, 1430,-

The "-" means I didn't calculate the difference, I could see it was too big.

I originally got the prime factorization by observing that:

Divisible by 10 therefore, 2*5*429.
429 divisible by 3, or 9? (add the digits together 4+2+9=15, therefore 3 is a divisor, but not 9), so 3*143=429.
Factor of 143, I notices it looks kind of close to 144=12*12, so I looked closer, and confirmed 11 is a factor. Now I have all the prime factors.

Answer 3

The easiest way to do the factoring is to plug the numbers into a computer algebra system, which spits out the answer (n-22)(3n+65).

The second easiest way to do this is to use the quadratic formula to find the two roots of the equation. If your roots are r1 and r2, then your factors are k(n-r1)(n-r2), where k is the coefficient of the n^2 term. If one of the factors contins fractions, distributing k over that factor will often get rid of them and make your final result look nicer, but it is not required for correctness. I would never recommend trying to factor large numbers, or polynomials containing large numbers, manually unless you absolutely have to - it takes too darn long.

Answer 4

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Answer 5

My favorite way of doing this if i don't know how to factorize it, is by using the quadratic formula

x = (-b ± sqrt(b^2 - 4ac))/2a

3n^2 - n - 1430
b = -1
a = 3
c = -1430

n = (-(-1) ± sqrt((-1)^2 - 4(3)(-1430)))/(2(3))
n = (1 ± sqrt(1 + 17160))/6
n = (1 ± sqrt(17161))/6
n = (1 ± 131/6)
n = (132/6) or (-130/6)
n = 22 or (-65/3)

When you factor this, you get

(3x + 65)(x - 22)

Answer 6

Look at the second page of the lesson at

Answer 7

The quadratic formula will always find the two solutions.

Answer 8

well, its hard try, 6_*=2/4=+9 try that