SOLUTION: Hello! A polynomial P is given: P(x) = 2x^3 + 7x^2 + 4x - 4 Find all the real zeros of P thanks for the homework help!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello! A polynomial P is given: P(x) = 2x^3 + 7x^2 + 4x - 4 Find all the real zeros of P thanks for the homework help!      Log On


   



Question 199776: Hello!
A polynomial P is given:
P(x) = 2x^3 + 7x^2 + 4x - 4
Find all the real zeros of P
thanks for the homework help!

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, let's find the possible rational roots.


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 -4 (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



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


Now let's test the possible roots to see if they are actually roots.


Let's see if the possible zero 1 is really a root for the function 2x%5E3%2B7x%5E2%2B4x-4


So let's make the synthetic division table for the function 2x%5E3%2B7x%5E2%2B4x-4 given the possible zero 1:
1|274-4
| 2913
29139

Since the remainder 9 (the right most entry in the last row) is not equal to zero, this means that 1 is not a zero of 2x%5E3%2B7x%5E2%2B4x-4


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


Let's see if the possible zero 1%2F2 is really a root for the function 2x%5E3%2B7x%5E2%2B4x-4


So let's make the synthetic division table for the function 2x%5E3%2B7x%5E2%2B4x-4 given the possible zero 1%2F2:
1/2|274-4
| 144
2880

Since the remainder 0 (the right most entry in the last row) is equal to zero, this means that 1%2F2 is a zero of 2x%5E3%2B7x%5E2%2B4x-4



So this means that 2x%5E3%2B7x%5E2%2B4x-4=%282x-1%29%28x%5E2%2B4x%2B4%29


Note: the quotient of the division results from taking half of the first three values in the bottom row.


Note: if you didn't find a root, then you would have to keep going until either you find one or you are done with the list.


Now let's solve x%5E2%2B4x%2B4=0 to find the next two zeros:



x%5E2%2B4x%2B4=0 Start with the given equation.


Notice we have a quadratic in the form of Ax%5E2%2BBx%2BC where A=1, B=4, and C=4


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-%284%29+%2B-+sqrt%28+%284%29%5E2-4%281%29%284%29+%29%29%2F%282%281%29%29 Plug in A=1, B=4, and C=4


x+=+%28-4+%2B-+sqrt%28+16-4%281%29%284%29+%29%29%2F%282%281%29%29 Square 4 to get 16.


x+=+%28-4+%2B-+sqrt%28+16-16+%29%29%2F%282%281%29%29 Multiply 4%281%29%284%29 to get 16


x+=+%28-4+%2B-+sqrt%28+0+%29%29%2F%282%281%29%29 Subtract 16 from 16 to get 0


x+=+%28-4+%2B-+sqrt%28+0+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-4+%2B-+0%29%2F%282%29 Take the square root of 0 to get 0.


x+=+%28-4+%2B+0%29%2F%282%29 or x+=+%28-4+-+0%29%2F%282%29 Break up the expression.


x+=+%28-4%29%2F%282%29 or x+=++%28-4%29%2F%282%29 Combine like terms.


x+=+-2 or x+=+-2 Simplify.


So the two solutions are x+=+-2 or x+=+-2


We can simply write this as x=-2 with a multiplicity of 2


===========================================================================

Answer:

So the three zeros of P%28x%29+=+2x%5E3+%2B+7x%5E2+%2B+4x+-+4 are: x=1%2F2 and x+=+-2 (with a multiplicity of 2)


If you have any questions, email me at jim_thompson5910@hotmail.com