# Third degree polynomial equations

• Feb 27th 2009, 10:22 PM
champrock
Third degree polynomial equations
Hi

I am stuck on a problem :

[ x^3 + 7 ] / [x^2 + 1 ] = 5

Question is : How many real roots does this equation have?

IN this I cant use a graphing calculator. Nor do i have access to even a simple calculator.

I think this can be done using the Intermediate value theorm (if the value of the function is positive and negative for some values of x then it must pass through X axis because it is continous.) But this does not seem a very scientific method to me. Moreover, this is hit and try.

Any "proper" method to solve such questions?
• Feb 27th 2009, 10:33 PM
mr fantastic
Quote:

Originally Posted by champrock
Hi

I am stuck on a problem :

[ x^3 + 7 ] / [x^2 + 1 ] = 5

Question is : How many real roots does this equation have?

IN this I cant use a graphing calculator. Nor do i have access to even a simple calculator.

I think this can be done using the Intermediate value theorm (if the value of the function is positive and negative for some values of x then it must pass through X axis because it is continous.) But this does not seem a very scientific method to me. Moreover, this is hit and try.

Any "proper" method to solve such questions?

Not hard to see (and without calculator) from the IVT that $x^3 -5x^2 + 2$ has roots in the intervals (-1, 0), (0, 1) and (4, 5).
• Feb 28th 2009, 07:29 AM
champrock
how did u get those intervals?

What i had done was that I had differentiated the function $x^3 -5x^2 + 2$ and got the points where slope is zero as x=10/3 and x=0.

At x=0 the function takes the value 2
and at x=10/3 the function takes the value -16.518.

So, there must exist atleast one real root where the functions value is 0.

Any input will be appreciated.
• Feb 28th 2009, 08:02 AM
e^(i*pi)
Quote:

Originally Posted by champrock
how did u get those intervals?

What i had done was that I had differentiated the function $x^3 -5x^2 + 2$ and got the points where slope is zero as x=10/3 and x=0.

At x=0 the function takes the value 2
and at x=10/3 the function takes the value -16.518.

So, there must exist atleast one real root where the functions value is 0.

Any input will be appreciated.

I don't think you need to differentiate:

$x^3 + 7 = 5x^2 +5$

and so $x^3 -5x^2 + 2 = 0$

and then follow on from mr fantastic
• Feb 28th 2009, 10:05 AM
Abu-Khalil
If you cannot see it easily, use the [HTML]http://en.wikipedia.org/wiki/Synthetic_division[/HTML]
• Feb 28th 2009, 01:27 PM
mr fantastic
Quote:

Originally Posted by champrock
how did u get those intervals?

What i had done was that I had differentiated the function $x^3 -5x^2 + 2$ and got the points where slope is zero as x=10/3 and x=0.

At x=0 the function takes the value 2
and at x=10/3 the function takes the value -16.518.

So, there must exist atleast one real root where the functions value is 0.

Any input will be appreciated.

Trying the interval (4, 5) is suggested by the coefficient of x, since 5^3 = 5 (5^2) and so the cubic will equal 2.

The other intervals involve simple values of x and are easily seen by inspection.