Let f(x) = x² + 6x²-x+ 5 then ,
number to be added be P
then,
f(x) = x² + 6x²-x+ 5 +P
According to the qn,
(x+3) is exactly divisible by zero then,
R=0
comparing .. we get a= -3
now by remainder theorm
R=f(a)
0=f(-3)
0=(-3)² + 6(-3)²-(-3)+ 5 + P
0= 9 + 54 + 3 + 5 + P
-71=P
therefore, -71 should be added.
Hope you understand
Answer:
its 11/6
Step-by-step explanation:
GCD of 11 and 6 is 1
Divide both the numerator and denominator by the GCD
11 ÷ 1
6 ÷ 1
Reduced fraction:
11/6
Answer:
how sid jame even get 64p bacteria cells like bro that cant be on exact he'd be 1000 years old if he knew exactly how much there was
Answer:
Linear
Dk if your asking for all or not