Question

Create your own third degree polynomial that when divided by x + 2 has a remainder of –4. Please Help

Answer 1

\[\it can be a \left( ax ^{2}+bx+c \right)\left( x+2 \right)-3,where a \neq 0\]
a,b,c are integers. you can plug any value of a,b,c except a=0

Answer 2

Dude Im the dumbest kid when it comes to math, can u please narrow it down for me?

Answer 3

\[\left( 2x ^{2}+3x+4 \right)\left( x+2 \right)-3=2x ^{3}+3x ^{2}+4x+4x ^{2}+6x+8-3\]
\[=2x ^{3}+7x ^{2}+10x+5\]

Answer 4

sure that it has remainder of -4?

Answer 5

i got a remainder of -3 for that polinimial

Answer 6

@julian25 yeah its the question they asked me on my online class bro!

Answer 7

ok man, thanks!!

Answer 8

here remainder is -3 ,for -4 replace-3 by -4

Answer 9

Okay, so what exactly do I put on my paper guys? :$

Answer 10

maybe the problem could be easy
when the problem says
divide the polinomial by (x-2) and get a value of -4
we know by the remainder theorem that
f(-2) =-4

Answer 11

it could be any polinomial of third degree
for example
f(x) =ax^3

Answer 12

replacing
-4 = a(-2)^3
-4 =-8a
a=1/2
so a possible polinomial is
f(x) = (1/2)x^3

Answer 13

ok so thats what i put on my paper?

Answer 14

jajajjaja lol
i dont understodd right

Answer 15

u*

Answer 16

one posible polinomial is (1/2)x^3
there is another answers but that polinomial is right