How do you factor completely #6c^2-33c+15#?

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Answer 1 · 2016-04-29T15:01:19

# (6c - 3) ( c - 5) # is the factorised form of the expression.

#6c^2 - 33c + 15#

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like #xc^2 + yc + z#, we need to think of 2 numbers such that:

#N_1*N_2 = x * z = 6 * 15 = 90#

AND

#N_1 +N_2 = y = -33#

After trying out a few numbers we get #N_1 = -30# and #N_2 =-3#
#(-30 )* (-3) = 90 #, and #(-30 )+( - 3)= -33#

#6c^2 - 33c + 15 = 6c^2 - 30c - 3c + 15#

# = 6c ( c - 5) - 3 ( c +5)#

# = (6c - 3) ( c - 5) #