Is it possible to factor #y= 3x^2+5x+2 #? If so, what are the factors?

1 Answer
iceman
Jan 1, 2016

#y=(3x+2)(x+1)#

Explanation:

#y=3x^2+5x+2#

A quadratic function has a general form of: #y=ax^2+bx+c#
when the leading coefficient(a) is not 1, we use an extension of factoring by grouping called the ac method as follows:

1) Multiply together ac and list the factors of ac
2) Find a pair that adds to b, If you can't find such pair the quadratic is a prime and does not factor.
3) Rewrite the middle term as a sum of terms whose coefficients are the chosen pair.
4) Factor by grouping.
Remember you should always first pull out the Greatest Common Factor GCF if it exists.

Now in this case:
#y=3x^2+5x+2# => ac = 3*2=6, pair is(3, 2) since 3 + 2 = 5
#y=3x^2+3x+2x+2# => factor by grouping:
#y=3x(x + 1) + 2(x + 1)# => complete the factoring:
#y=(3x+2)(x+1)#

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