SOLUTION: i am completely lost as to how to even start this problem. Can anyone help me? find the zeros of (2x-7)^2(x^2-9)and state the multiplicity of each. thanks so much!!!!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: i am completely lost as to how to even start this problem. Can anyone help me? find the zeros of (2x-7)^2(x^2-9)and state the multiplicity of each. thanks so much!!!!      Log On


   

Question 107107: i am completely lost as to how to even start this problem. Can anyone help me? find the zeros of (2x-7)^2(x^2-9)and state the multiplicity of each. thanks so much!!!!
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
find the zeros of
%282x-7%29%5E2%28x%5E2-9%29
First, you can write %282x-7%29%5E2 as %282x-7%29%282x-7%29
And %28x%5E2-9%29 as+%28x-3%29%28x%2B3%29
Then you will have:
%282x-7%29%282x-7%29+%28x-3%29%28x%2B3%29
Zeros: product is equal to 0 if one factor is equal to 0
So,
2x-7=0……… => …2x+=+7… => x+=+7%2F2....first and secon zero ( we have 2x+-7 two times
Or
x-3+=+0……… =>…x+=+3....third zero
Or
x+%2B+3+=+0……..=> …x+=+-3....fourth zero
check:
%282x-7%29%5E2%28x%5E2-9%29
plug in zeros:
if x=7%2F2
(since one factor is 0
Or
x=3
%282%2A3-7%29%5E2%283%5E2-9%29+=+%286-7%29%5E2%289-9%29+=1%2A0+=+0
Or
x+=+-3
%282%2A%28-3%29-7%29%5E2%28%28-3%29%5E2-9%29+=+%28-6-7%29%5E2%289-9%29+=1%2A0+=+0