SOLUTION: (22 pts) Consider the polynomial f(x) = 2x^3 – 3x^2 – 8x – 3. (i) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.

Algebra ->  Rational-functions -> SOLUTION: (22 pts) Consider the polynomial f(x) = 2x^3 – 3x^2 – 8x – 3. (i) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.       Log On


   

Question 148299: (22 pts) Consider the polynomial f(x) = 2x^3 – 3x^2 – 8x – 3.
(i) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.








(ii) Find all of the zeros of the given polynomial. Be sure to show work, explaining how you have found them.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
i)

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 2 (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







--------------------------------------------------------


ii)

With the help of a graphing calculator, we see that -1 is a zero of 2x%5E3-3x%5E2-8x-3

note: let me know if you need to find the zeros a different way.

So let's set up a synthetic division table by placing the value -1 in the upper left corner and placing the coefficients of the polynomial to the right of -1.
-1|2-3-8-3
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
-1|2-3-8-3
|
2

Multiply -1 by 2 and place the product (which is -2) right underneath the second coefficient (which is -3)
-1|2-3-8-3
|-2
2

Add -2 and -3 to get -5. Place the sum right underneath -2.
-1|2-3-8-3
|-2
2-5

Multiply -1 by -5 and place the product (which is 5) right underneath the third coefficient (which is -8)
-1|2-3-8-3
|-25
2-5

Add 5 and -8 to get -3. Place the sum right underneath 5.
-1|2-3-8-3
|-25
2-5-3

Multiply -1 by -3 and place the product (which is 3) right underneath the fourth coefficient (which is -3)
-1|2-3-8-3
|-253
2-5-3

Add 3 and -3 to get 0. Place the sum right underneath 3.
-1|2-3-8-3
|-253
2-5-30

Since the last column adds to zero, this means that -1 is a zero of 2x%5E3-3x%5E2-8x-3 (this confirms our original claim).



Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,-5,-3) form the quotient

2x%5E2+-+5x+-+3


So %282x%5E3+-+3x%5E2+-+8x+-+3%29%2F%28x%2B1%29=2x%5E2+-+5x+-+3


Basically 2x%5E3+-+3x%5E2+-+8x+-+3 factors to %28x%2B1%29%282x%5E2+-+5x+-+3%29


Now lets find the zeros for 2x%5E2+-+5x+-+3.


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%282%29%28-3%29+%29%29%2F%282%282%29%29 Plug in a=2, b=-5, and c=-3


x+=+%285+%2B-+sqrt%28+%28-5%29%5E2-4%282%29%28-3%29+%29%29%2F%282%282%29%29 Negate -5 to get 5.


x+=+%285+%2B-+sqrt%28+25-4%282%29%28-3%29+%29%29%2F%282%282%29%29 Square -5 to get 25.


x+=+%285+%2B-+sqrt%28+25--24+%29%29%2F%282%282%29%29 Multiply 4%282%29%28-3%29 to get -24


x+=+%285+%2B-+sqrt%28+25%2B24+%29%29%2F%282%282%29%29 Rewrite sqrt%2825--24%29 as sqrt%2825%2B24%29


x+=+%285+%2B-+sqrt%28+49+%29%29%2F%282%282%29%29 Add 25 to 24 to get 49


x+=+%285+%2B-+sqrt%28+49+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


x+=+%285+%2B-+7%29%2F%284%29 Take the square root of 49 to get 7.


x+=+%285+%2B+7%29%2F%284%29 or x+=+%285+-+7%29%2F%284%29 Break up the expression.


x+=+%2812%29%2F%284%29 or x+=++%28-2%29%2F%284%29 Combine like terms.


x+=+3 or x+=+-1%2F2 Simplify.


So the zeros of 2x%5E3-3x%5E2-8x-3 are x=-1, x=3, or x=-1%2F2