SOLUTION: Find all the zeros of f (x) = 12x4 — 67x3 + 108x2 — 47x + 6

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Question 113804: Find all the zeros of f (x) = 12x4 — 67x3 + 108x2 — 47x + 6
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
From part C), we found x=2 to be a zero for 12x%5E4-67x%5E3%2B108x%5E2-47x%2B6 (ie f(2)=0) and the quotient was 12x%5E3+-+43x%5E2+%2B+22x+-+3. So let's find the zeros for 12x%5E3+-+43x%5E2+%2B+22x+-+3

For 12x%5E3+-+43x%5E2+%2B+22x+-+3, the same sign rules apply as they do to the function 12x%5E4-67x%5E3%2B108x%5E2-47x%2B6: there are no negative zeros and there are 4, 2, or no positive zeros


So let's find the possible zeros for 12x%5E3+-+43x%5E2+%2B+22x+-+3

Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of -3 (the last coefficient):



Now let's list the factors of 12 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur





To save time, I'm only going to use synthetic division on the possible positive zeros (using Descartes rule of signs) that are actually zeros of the function.
Otherwise, I would have to use synthetic division on every possible positive root (there are 12 possible positive roots, so that means there would be at most 12 synthetic division tables).
However, you might be required to follow this procedure, so this is why I'm showing you how to set up a problem like this


If you're not required to follow this procedure, simply use a graphing calculator to find the roots




So let's use the zero x=3 (which is an actual zero)



Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
3|12-4322-3
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 12)
3|12-4322-3
|
12

Multiply 3 by 12 and place the product (which is 36) right underneath the second coefficient (which is -43)
3|12-4322-3
|36
12

Add 36 and -43 to get -7. Place the sum right underneath 36.
3|12-4322-3
|36
12-7

Multiply 3 by -7 and place the product (which is -21) right underneath the third coefficient (which is 22)
3|12-4322-3
|36-21
12-7

Add -21 and 22 to get 1. Place the sum right underneath -21.
3|12-4322-3
|36-21
12-71

Multiply 3 by 1 and place the product (which is 3) right underneath the fourth coefficient (which is -3)
3|12-4322-3
|36-213
12-71

Add 3 and -3 to get 0. Place the sum right underneath 3.
3|12-4322-3
|36-213
12-710

Since the last column adds to zero, we have a remainder of zero. This means x-3 is a factor of 12x%5E3+-+43x%5E2+%2B+22x+-+3

Now lets look at the bottom row of coefficients:

The first 3 coefficients (12,-7,1) form the quotient

12x%5E2+-+7x+%2B+1


So %2812x%5E3+-+43x%5E2+%2B+22x+-+3%29%2F%28x-3%29=12x%5E2+-+7x+%2B+1

You can use this online polynomial division calculator to check your work

Basically 12x%5E3+-+43x%5E2+%2B+22x+-+3 factors to %28x-3%29%2812x%5E2+-+7x+%2B+1%29

Now lets break 12x%5E2+-+7x+%2B+1 down further

12x%5E2+-+7x+%2B+1=0 Set the expression equal to zero and solve for x


Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve 12%2Ax%5E2-7%2Ax%2B1=0 ( notice a=12, b=-7, and c=1)




x+=+%28--7+%2B-+sqrt%28+%28-7%29%5E2-4%2A12%2A1+%29%29%2F%282%2A12%29 Plug in a=12, b=-7, and c=1



x+=+%287+%2B-+sqrt%28+%28-7%29%5E2-4%2A12%2A1+%29%29%2F%282%2A12%29 Negate -7 to get 7



x+=+%287+%2B-+sqrt%28+49-4%2A12%2A1+%29%29%2F%282%2A12%29 Square -7 to get 49 (note: remember when you square -7, you must square the negative as well. This is because %28-7%29%5E2=-7%2A-7=49.)



x+=+%287+%2B-+sqrt%28+49%2B-48+%29%29%2F%282%2A12%29 Multiply -4%2A1%2A12 to get -48



x+=+%287+%2B-+sqrt%28+1+%29%29%2F%282%2A12%29 Combine like terms in the radicand (everything under the square root)



x+=+%287+%2B-+1%29%2F%282%2A12%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%287+%2B-+1%29%2F24 Multiply 2 and 12 to get 24

So now the expression breaks down into two parts

x+=+%287+%2B+1%29%2F24 or x+=+%287+-+1%29%2F24

Lets look at the first part:

x=%287+%2B+1%29%2F24

x=8%2F24 Add the terms in the numerator
x=1%2F3 Divide

So one answer is
x=1%2F3



Now lets look at the second part:

x=%287+-+1%29%2F24

x=6%2F24 Subtract the terms in the numerator
x=1%2F4 Divide

So another answer is
x=1%2F4

So the solutions for 12%2Ax%5E2-7%2Ax%2B1 are:
x=1%2F3 or x=1%2F4




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Answer:


So the zeros are x=1%2F3, x=1%2F4, x=2, x=3