How do you find #(g/h)(a)# given #g(a)=-3a^2-a# and #h(a)=-2a-4#?

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Answer 1 · 2018-01-26T23:04:47

See a solution process below:

#(g/h)(a) = (g(a))/(h(a)) = (-3a^2 - a)/(-2a - 4)#

We can remove the negatives by multiplying the result by the appropriate form of #1#:

#(g/h)(a) = (-1)/-1 xx (-3a^2 - a)/(-2a - 4)#

#(g/h)(a) = (-1(-3a^2 - a))/(-1(-2a - 4))#

#(g/h)(a) = (3a^2 + a)/(2a + 4)#

If necessary, we can factor the numerator and denominator as:

#(g/h)(a) = (a(3a + 1))/(2(a + 2))#