Factor something without quadratic formula?

Question

I'm supposed to be factoring without the quadratic formula. One of the problems is:

7y^2 - 21y +5

Answer 1

There's a positive term on the end and a negative term in the middle, so we want something in the form of
(?x + ?)(?x - ?)

The only factors of 7 are 1 and 7, so this means we have
(7x + ?)(x - ?), or (x + ?)(7x - ?)

The only factors of 5 are 1 and 5, so this means the only possible answers are
(7x + 1)(x - 5), (7x + 5)(x - 1), (7x - 1)(x + 5), (7x - 5)(x + 1)

However, none of these combinations gives you a middle term of -21y, so you can't factor it. At least not with integers.

I suppose you could set the expression equal to zero, solve via completing the square, and then use those solutions in your factors, bringing the 7 back.

Answer 2

The answer is 7 (y - 3/2 + sqrt( 43/28)) (y - 3/2 - sqrt( 43/28)).

Start with:
7y^2 - 21y + 5 = 0

Divide by 7:
y^2 - 3y + 5/7 = 0

Move the constant to the right side
y^2 - 3y = - 5/7

Complete the square
y^2 -3y + 9/4 = - 5/7 + 9/4
The "9" comes from 3^2, and then you divide by 4.

(y - 3/2)^2 = -20/28 + 63/28 = 43 / 28

Square root of each side:
(y - 3/2) = [plus or minus] sqrt( 43/28 )

Add 3/2 to each side
y = 3/2 [plus or minus] sqrt( 43/28 )

The two main factors are then
(y - 3/2 + sqrt( 43/28)
and
(y - 3/2 - sqrt( 43/28)

Also you need to put the 7 back in front.

Answer 3

This is your problem

(ay +b ) (cy +d)

ac = 7
bd = 5

Finally, ad + bc = -21

Since 7 and 5 are prime, it would seem that a,b,c and d are not integers. Good luck. You can't make this the difference of two squares.

Answer 4

are you sure you didn't write the problem on here wrong
the problem you wrote will not factor

but if the problem is 7y^2-12y+5
it will factor
(-7y+5)(-y+1)=
7y^2-7y-5y+5=
7y^2-12y+5