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:
Step-by-step explanation:
Convert the equation of a circle in general form shown below into standard form. Find the center and radius of the circle. Group the x 's and y 's together. Consider the x2 and x terms only. Complete the square on these terms. Replace the x2 and x terms with a squared bracket.
I believe the answer is a.
Use the Quadratic formula,
x= -b +/- sqrt(b^2-4ac)
-------------------------
2a
x= 4 +/- sqrt(16+12)
----------------------
-6
x = 4+/- sqrt(28)
---------------
-6
x = 4 +/- 2*sqrt(7)
------------------
-6
x= 2(2+/- sqrt(7))
------------------
2(-3)
x= 2+/- sqrt(7)
--------------
-3
The answer to your question is E
