[jacks]

Let \( R(x) \) be remainder when

\( P(x)=x^{73}-2x^6+3x^3 \) is divided by

\( x^4(x^2-1) \).Then the number of real

roots of the equation \( R''(x)=0 \), is

[Luis Fuentes]

Hi

Let \( R(x) \) be remainder when

\( P(x)=x^{73}-2x^6+3x^3 \) is divided by

\( x^4(x^2-1) \).Then the number of real

roots of the equation \( R''(x)=0 \), is

We have:

\( x^{73}-2x^6+3x^3=x^4(x^2-1)q(x)+R(x) \) (*)

where \( q(x) \) is  the quotient and \( R(x) \) is the remiander. From (*) we deduce that \( x^3 \) divides \( R(x) \) so \( R(x)=x^3f(x) \) where \( f(x) \) is a second-degree polynomial and:

\( x^{70}-2x^3+3=x(x^2-1)q(x)+f(x) \)

Evalualting at \( x=-1,x=0,x=1 \):

\( 6=f(-1),\quad 3=f(0),\quad 2=f(1) \)

Since \( f(x) \) is of degree two, we can  determine it if we know its value at three points. We obtain:

\( f(x)=x^2-2x+3 \)

and

\( R(x)=x^3(x^2-2x+3) \), \( R'(x)=5x^4-8x^3+9x^2 \), \( R''(x)=20x^4-24x^2+18x=2x(10x^2-12x+9) \)

The discriminant of \( 10x^2-12x+9=0 \) is \( 12^2-4\cdot 9\cdot 10<0 \), so it has not real roots.

The unique real root of \( R''(x) \) is \( x=0 \).

Best regards.

[jacks]

Thanks Admin Got it.

\( \displaystyle x^{73}-2x^6+3x^3=x^{67}(x^6-x^4)+x^{65}(x^6-x^4)+x^{63}(x^6-x^4)+x^{61}(x^6-x^4)+x^3(x^6-x^4)+x(x^6-x^4)\cdots -2(x^6-x^4)+x^5-2x^4+3x^3 \)

So \( \displaystyle R(x)=x^5-2x^4+3x^3 \)

Then \( \displaystyle R''(x)=20x^3-24x+18x  \)

[Richard R Richard]

Then \( \displaystyle R''(x)=20x^3-24x+18x  \)

Hello, you have a typo
\( \displaystyle R''(x)=20x^3-24x^2+18x  \)



\( \displaystyle R''(x)=2x(10x^2-12x+9)  \) then $$x=0$$ is one root


$$0=10x^2-12x+9$$


It has no real roots as Luis previously told you.