How do you factor #6y^2-5y+1#?

1 Answer
Aug 7, 2015

Answer:

#(2y-1)(3y-1)#

Explanation:

When factorizing a quadratic function look for factors of the square coefficient (in this case #6#) and factors of the constant term (in this case #1#).

In this case the coefficient of the #y# term is negative and the constant term is positive, so the constants in each of the factors must both be negative. Since the constant is #+1#, its factors must be #-1# and #-1#.

Integer factors of #6#, independent of order, are

#(1,6); (2,3); ("-"1,"-"6); ("-"2,"-"3)#

Now look for the factor pair that when multiplied by #(-1, -1)# sums to #-5#. Here this is #(2,3)#. These are the coefficients of the #y# terms in each factor.

Hence, the factors are #(2y-1)# and #(3y-1)#.

Note: Not all quadratic functions can be factorized.