How would I factor these trinomials?

Question

1.) 3p^2+p-10
2.) 16y^2-6y-1
3.) 6x^2-x-12
4.) 10n^2-n-2

Answer 1

(3p-5)(p+2)
(8y+1)(2y-1)
(2x-3)(3x+4)
(2n-1)(5n+2)

Answer 2

Use this methodology:
Take the trinomial of the form ap^2+bp+c and set it equal to 0. Solve for the roots using the quadratic formula -b/2a +- sqrt(b^2-4ac)/2a. The factorization is of the form (p - root1)(p - root2)

Ex. 3p^2 + p -10 = 0
roots = -1/6 +- sqrt(1-4*3*-10)/6 = -1/6 +- 11/6
root1 = -1/6 -11/6 = -2
root2 = -1/6 +11/6 = 10/6 = 5/3
This solves (p + 2)(p - 5/3) = 0
The factorization is (p + 2)(p - 5/3)

Answer 3

The rule is:

3p^2+p-10
___*___=-30 (a*c)
___+___=1(b)

find those 2 answers and stick em in...

3p^2-6p+5p-10

Then split em up, and factor as much as you can

3p(p-2) + 5(p-2)

Notice the similarity, this has to be there otherwise its wrong (p-2)

so you got:

(3p+5)(p-2) =0

3p=-5

SO........

p= -5/3 and p=2

Follow those steps for all of them to get the answers.. :)

Answer 4

one way is to use the quadratic formula (if you don't know it, just look it up, it's all over the web) to find the roots (where the variable equals zero) and work backwards

basically, if x=a/b, then the factor is (bx-a)

so if x=3/4, the factor is (4x-3)
if x=-9, the factor is (x+9)

Answer 5

I think you have to complete the square or something weird like that