Question

Find factors of 15y^2-50y+35=

Answer 1

I would first factor out that 5: 15y^2-50y+35 = 5(3y^2 - 10y + 7)
and then focus on factoring 3y^2 - 10y + 7. Looking at the first and last coefficients 3 and 7, I'd form possible rational roots, which would include 7/3 and -7/3.

Using synthetic division, I'd determine whether either 7/3 or -7/3 is a root of 3y^2 - 10y + 7:

____________
(7/3) / 3 -10 7
7 -7
--------------------
3 -3 0

Since there is no remainder, I'd conclude that 7/3 is a root of 3y^2 - 10y + 7, and that (3x-7) is a factor. Look at the coefficients 3 -3; they tell us that (3x-3) is another factor.

Now let's take back that '5' we factored out earlier.

Factors of 15y^2 - 50y + 35 are 5, (3x-7) and (x-1).

Multiply these 3 factors together as a check. I found that (3x-7)(x-1) comes out to 3x^2 - 10x + 7, and that multiplying this by 5 produces

15y^2 - 50y + 35, which is the expression we started with.

Answer 2

First, factor out the common factor 5.

15y^2-50y+35 =

= 5(3y^2 - 10y + 7)

Now we need to factor the trinomial.
3 can be factored as 3 * 1
7 can be factored as 1 * 7 or (-1) * (-7)
Since the middle term is negative, we need to use -1 and -7 as the factors of 7

= 5(3y - 7)(y - 1)