how can you factor..?

Question

how can you factor this 2a^2 - 13a + 15 ?

Answer 1

OK, the answer will look like:

(2a - n1)(a-n2) since the middle term is negative and the constant factor is positive. The number 15 can be factored as (1,15) or (5,3). By playing around with these pairs, you eventually get

(2a- 3)(a-5)

there it is!

Answer 2

2a^2 - 13a + 15

Hopefully there are numbers b,c,d, and e, such that
(ba + c)(da + e) = 2a^2 - 13a + 15

You know that bd = 2. That pretty much means that either b or d is 2 and the other of them is 1. And if one is negative, they're both negative otherwise ab would be -2

You also know that ce = 15. There are a couple of possibilites here. One of them could be 15 and the other 1, or one is 5 and the other is 3. And again, you know they are both positive or they are both negative, or ce would be -15.

You also know that be+cd = -13.
Possibility 1: b=2, e= -5, c=-3, d= 1
(2)(-5) + (-3)(1) = -13
or Possibility 2: b=2, e=1, c=-15, d=1
and (2)(1) + (-15)(1) = -13.
But we can eliminate the second guess because (1)(-15) = -15 instead of +15.
So the factors have to be based on the first possibility.
(2a - 3)(a - 5)

You check your factors by multiplying them out.
(2a - 3)(a - 5)=
(2a - 3)(a) + (2a - 3)(-5) Because of the distributive property of multiplication over addition. (a)(b+c)=(ab)+(ac)
= (a)(2a - 3) + (-5)(2a - 3) Because of the commutative property of multiplication ab=ba
= [2a^2 + (-3)(a)] + [(-5)(2)(a) + (-5)(-3)] Because of distributivity again.
= 2a^2 + [(-3)(a) + (-10)(a)] + 15 Because of the associative property of addition. (a+b)+c=a+(b+c), and a little arithmetic.
= 2a^2 + [(-3) + (-10)]a + 15 Distributivity again.
= 2a^2 + (-13)a + 15 Which can also be written
2a^2 - 13a + 15

Answer 3

ok set it up like

(2a )(a )

now find factorst of 15. 5 and 3

(2a-3)(a-5)

Answer 4

(2a-3)(a-5)
a=3/2 or 5

Answer 5

(2a - 3) (a - 5)
you can also use the quadratic formula for this or similar problems that are unfactorable: go here to get the formula: http://www.sosmath.com/algebra/quadraticeq/quadraformula/quadraformula.html

Answer 6

(2a - 3) (a - 5)

Answer 7

remember: ax^2+bx+c
to factorize: [(-b) +/- square root of((b^2)-4ac)] / 2a

so,
[(13)+/- square root of ((-13)^2-4(2)(15))] / 2(2)
[13 +/- square root of (169-120)] / 4
[13 +/- square root of 49] / 4
[13 +/- 7] / 4
1st factor: (13+7)/4 = 5
2nd factor: (13-7)/4 = 3/2

:::try to practice this with other problems:::
good luck ;)

Answer 8

(2a-3)(a-5)

Answer 9

(2a-3)(a-5)

Answer 10

(2a-3)*(a-5)