Algebra II with Trig Ch1.4 #25

by | Jan 29, 2016

Question:

We need help on Algebra II with Trig 1.4 #25. How do we get started?

Answer:

We want to reduce the problem.

The first step I take on these kind of problems is to make it easier to see. Looking at what we are given – we can see that the we have two big parts added together in the numerator

One part is the

-2x(x+4)^3

and the other part is the

-3(3-x^2)(x+4)^2

We can see that both parts have an x+4.

Often in busy problems like this, we can make things easier and replace the x+4 with k to get something like

-2xk^3 – 3(3-x^2)k^k

as the numerator. So now we can see we can factor out the k.

k^2 [-2xk – 3(2-x^2)]

We can put this back to get

(x+4)^2 [-2x(x+4) – 3(2-x^2)]  /  [ (x+4)^6]

Now we can reduce by dividing out the (x+4)^2 term to get

[-2x(x+4) – 3(2-x^2)]  /  [ (x+4)^2]

Written By DaleCallahan

Related Posts

0