Question

The zeroes of the polynomial
P(2) ax² + bxtc
are 1 and 2
Find the value of a, b and c​

Answer 1

Answer:

You have to solve the system

{p(1+3–√)=0p(2)=−2

{a(1+3–√)2+b(1+3–√)+c=04a+2b+c=−2

{a(4+23–√)+b(1+3–√)+c=04a+2b+c=−2

Since a,b,c are rationals the first equation "splits" in the first two equations of the following system:

⎧⎩⎨⎪⎪4a+b+c=02a+b=04a+2b+c=−2