Question

q.no of intergral root of equation x^8-24x^7-18x^5+39x^2+1155​

Answer 1

Answer:

Step-by-step explanation:

The term a1x is missing from the polynomial f(x)=x8−24x7−18x5+39x2+1155 and 1155

is a squarefree integer; therefore (since ±1 are not solutions) the equation f(x)=0 does not have integer solutions.

For, if r∈Z is a solution then

r8−24r7−18r5+39r2+1155=0⇒r2(r6−24r5−18r3+39)=−1155⇒r2∣1155.

More generally if a0∈Z is not divided by a kth power of any prime number, then the only possible integer solutions of the equation

anxn+an−1xn−1+…+akxk+a0=0

are ±1.