Question

3y^2 - 13y-10 - Factoring Trinomials where a does not equal 1

Answer 1

\[(y-5) (3 y+2) \]

Answer 2

there is a method; do you want to learn

Answer 3

there and is trial and error

Answer 4

Yes please. I would like to learn the method.

Answer 5

if there are factors of ac that sum to b, then the trinomial is factorable\[ax^2+bx+c\]in this case ac=3*(-10)=130 and the factors are
1; 30
2; 15
3; 10
5; 6
where each pair of factors must have differing signs.
(continued)

Answer 6

the pair -15*2=-30 and -15+2=-13

so we can rewrite the -13y as -15y+2y to get\[3y^3-15y+2y-10\]
(continued)

Answer 7

now there is a method of factoring called "factoring by grouping" where you factor out the GCF of the first two terms and the GCF out of the last two terms:\[3y(y-5)+2(y-5)\](continued)
NOTE: I accidentally wrote y^3 instead of y^2 for the first term in the previous post!!

Answer 8

now id you think of this expression as having two terms each with a GCF of the binomial y-5, you can factor it out, leaving the 3y and the +2:\[(y-5)(3y+2)\]and this is the factored form

Answer 9

same as saifoo got using trial and error most likely which just takes some practice

Answer 10

Thank you for your answer. This will help me out greatly.

Answer 11

you want to do another; i've got one ready to do... with less explanation

Answer 12

Sure. By the way, i am new to this.

Answer 13

Factor \[6x^2+x-12\]

Answer 14

ac=6(-12)=-72
-8*9=-72
-8+9=1=b, thus\[6x^2-8x+9x-12\]factor out the GCF from each group\[=2x(3x-4)+3(3x-4)\]factor out the binomial\[=(3x-4)(2x+3)\]done