How would I factor the equation y2-7y+6=0?

y^2 - 7y + 6 = 0

(y-6)(y-1) = 0

Start from what you know and fill in the blanks:

y^2 - 7y + 6 = 0

Since the y^2 coefficient is 1, the coefficients of both y terms in the factors are ones:

(y +- __ ) * (y +- __ ) = 0


Since the constant is positive, the +- operator is the same in both factors. Since the y coefficient is negative, the operators must be minuses:

(y - __ ) * (y - __ ) = 0


All that's left is to find two factors of 6 which, when added, total 7. The values 1 and 6 do that:

(y - 1) * (y - 6) = 0

The two solutions of the equation are 1 and 6.

y^2 - 7y + 6 = 0
y^2 - y - 6y + 6 = 0
(y^2 - y) - (6y - 6) = 0
y(y - 1) - 6(y - 1) = 0
(y - 1)(y - 6) = 0

y - 1 = 0
y = 1

y - 6 = 0
y = 6

∴ y = 1 , 6