It’s A because I took this test
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:
Part A: 2k(2c2+5(2)-8c-20)
Part B: sorry i don't know how to do this part
Step-by-step explanation:
Part A, you divide each number by 2 first so you get a simplified version of each number. Then, you will quickly realize that the varible k is similar in all of the numbers then you remove that and put it with the 2 outside the (). Hope you understood my explination.
You only need to look at the y intercepts.
-11 - -7 is -4, but you can't have a negative distance so it's just 4
