SOLUTION: <img src="http://img338.imageshack.us/img338/9167/mathaq6.png"> I NEED HELP! :]

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Question 117516:
I NEED HELP! :]
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a%2Bb%2Bc%2Bd=0 Start with the given equation


d=-a-b-c Solve for d

ax%5E3%2Bbx%5E2%2Bcx-a-b-c=0 Plug in d=-a-b-c into ax%5E3%2Bbx%5E2%2Bcx%2Bd=0


%28ax%5E3-a%29%2B%28bx%5E2-b%29%2B%28cx-c%29=0 Group like terms


a%28x%5E3-1%29%2Bb%28x%5E2-1%29%2Bc%28x-1%29=0 Factor out the GCF from each individual group


a%28x-1%29%28x%5E2%2Bx%2B1%29%2Bb%28x%5E2-1%29%2Bc%28x-1%29=0 Factor x%5E3-1 by using the difference of cubes formula to get %28x-1%29%28x%5E2%2Bx%2B1%29


a%28x-1%29%28x%5E2%2Bx%2B1%29%2Bb%28x-1%29%28x%2B1%29%2Bc%28x-1%29=0 Factor x%5E2-1 by using the difference of squares formula to get %28x-1%29%28x%2B1%29


Notice how we have the common term x-1. We can factor this term out.


%28x-1%29%28a%28x%5E2%2Bx%2B1%29%2Bb%28x%2B1%29%2Bc%29=0 Factor out the GCF x-1


Now if we let H=x-1 and K=a%28x%5E2%2Bx%2B1%29%2Bb%28x%2B1%29%2Bc we'll get

H%2AK=0


So by the zero product property we get

H=0 or K=0


but since H=x-1, this means

x-1=0

Now solve for x


x=1


So x=1 is a zero of ax%5E3%2Bbx%5E2%2Bcx%2Bd=0 if the coefficients satisfy the equation a%2Bb%2Bc%2Bd=0



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Check:

Unfortunately, there is no formal check for this type of problem. But if we make sure that the coefficients satisfy the equation a%2Bb%2Bc%2Bd=0, then we can graph the equation and see that one root is x=1. For instance, let a=b=1 and c=d=-1 so we'll get


x%5E3%2Bx%5E2-x-1 (notice how 1%2B1-1-1=2-2=0 which satisfies the equation a%2Bb%2Bc%2Bd=0)


Now graph the polynomial to get


+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E3%2Bx%5E2-x-1%29+


and we can see that one root is x=1


If we try different values of a,b,c, and d that will satisfy the equation a%2Bb%2Bc%2Bd=0, we'll see that x=1 is a root every time. So in a way, our answer has been verified.