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:

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

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