How do you reduce #(w^2 + 5w+6)/(w^2-w-12)#?

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Answer 1 · 2016-04-15T15:31:45

#(w+2)/(w+4)#

Consider the numerator #(w^2+5w+6)#

Factors of 6 are: {1,6} ; {2,3}

The appropriate factors need to be such that we have +5 as the sum and +6 as the product.

The factors of {-1,6} give us 5 as a sum but not +6 as a product so it is not them

However:#" "(+2)xx(+3) =+6#

#" "+2+3=+5#

So this works, giving

#(w^2+5w+6)/(w^2-w-12) =((w+2)(w+3))/(w^2-w-12)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
applying the same approach to the denominator we end up with;

#((w+2)(w+3))/((w+3)(w+4)#

Write as:

#(w+2)/(w+4)xx(w+3)/(w+3)#

But #(w+3)/(w+3)=1# giving

#(w+2)/(w+4)#