Synthetic division/Rational Zeros Theorem

[ctess]

Ok here is my problem:

Given the polynomial f(x) = 2x^3 -5x^2- 4x + 3, find all the solutions if the function is completed.
a) f(x) =0
b) f(x+2)=0
d) f(2x) = 0

Ok I understand what to do for the first one: f(x) = 0

I found that 3, 1/2, and -1 were the zeros in this situation.

I will explain what I know so far and then continue on with what im confused about.

I know that we take the factors of the constant 3 which will = p ( rational zeros = p/q where p is the constant and q is teh leading coefficient) and q in this case is 2.

so the rational factors look like this: +/- 3, +/- 1, +/- 1/2 , +/- 3/2

Then you use synthetic division to see if there is a denominator if there isn't a denominator then that number is a zero for the function.

Ok now what im confused on is f(x+2) = 0.

Do I plug x+2 into the place of all the x's like this?

f(x) = 2(x+2)^3-5(x+2)^2-4(x+2)+3

If so that would result in:

f(x) = 2x^3-5x^2-4x-9

I have plugged in all the factors and none of them are zeros but my book says the zeros are: -3, -3/2, 1

So obviously im going wrong somewhere. any help would be much appreciated.

 

[skeeter]

whoa ... you are really making this too hard.

first off, you found the roots for f(x) = 0 ... x = 3, 1/2, -1 ... good job!

sounds simple, but recap what that means ... f(3) = 0, f(1/2) = 0, and f(-1) = 0.

for f(x+2) = 0 ... x+2 = 3, x+2 = 1/2, and x+2 = -1

so, the roots for f(x+2) = 0 are x = 1, x = -3/2, and x = -3

remember that f(x+2) is just f(x) shifted 2 units left ... note that all the roots shifted 2 units left also.

do the same for f(2x) = 0.

 

[ctess]

oh wow, so I can just take my original answers and shift them because the formulae doesn't change. I didn't even think about it like that.

wow thank you very much...You are right I did overly complicate it.