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:
1x+3x+4x= 128
8x=128
x=128/8
x=16
therefore,
room 1 = 1x
= 1*16=16
room 2 = 3x
=3*16=48
room 3 = 4x
=4*16=64
PLS MARK BRAINLIEST, THNX
Answer:
squares
Step-by-step explanation:
Hello,
Dividing by 0 is not defined so let s take x different from 4 and -4 and then we can write:
As
Thanks
