# factoring polynomials

G

#### Guest

##### Guest
I need help. It says factor 15y^2-30y+35

the answer I got was (5)(3y^2-6y+7) is it correct??

what i did was find a common factor for each and i came up with 5
and then I multiplied 5x3 for 15 then 5x6 for 30 and 5x7 for 35.... did i mess up?

G

#### Guest

##### Guest
I need help. It says factor 15y^2-30y+35

the answer I got was (5)(3y^2-6y+7) is it correct??

what i did was find a common factor for each and i came up with 5
and then I multiplied 5x3 for 15 then 5x6 for 30 and 5x7 for 35.... did i mess up?

15y² - 30y + 35

5 (3y² - 6y + 7)

No, I think you got it... Can someone verify whether or not a) this is correct, and b) if the inside cannot be factored any further (i.e., (y-x)²)?

#### skeeter

##### Senior Member
5(3y^2-6y+7) is as far as you can go ...

a quadratic of the form ax^2 + bx + c will factor if the discriminant (b^2 - 4ac) is a perfect square number.

3y^2 - 6y + 7

(-6)^2 - 4(3)(7) = 36 - 84 = -48 ... -48 is not a perfect square number.

G

#### Guest

##### Guest
skeeter said:
5(3y^2-6y+7) is as far as you can go ...

a quadratic of the form ax^2 + bx + c will factor if the discriminant (b^2 - 4ac) is a perfect square number.

3y^2 - 6y + 7

(-6)^2 - 4(3)(7) = 36 - 84 = -48 ... -48 is not a perfect square number.
Thank you for clarifying. I wasn't sure. Also, since you got a negative... if he had to solve for x, would he get no solution?

#### skeeter

##### Senior Member
if the quadratic 3y^2-6y+7 were set = 0, it would have two imaginary solutions.

G

#### Guest

##### Guest
skeeter said:
if the quadratic 3y^2-6y+7 were set = 0, it would have two imaginary solutions.

LOL! Funny you should say that... I was just explaining that to my father, who asked me what I was up to. I love math. :lol: