Question

Help me please!

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

My answer is f(x) = (x + 2)(5x^2 - 2x - 20) but im not sure i did it right. Can you explain to me what i did wrong if i did at all. And what i could change to make it better.

Answer 1

\[\left( x+2 \right)^{3}-4\]

Answer 2

Can you explain that a little?

Answer 3

when you divide by (x+2). the left part divides evenly and the raminder is -4. if you multiply out the cube and subtract 4 it will look like a regluar polynomial, but if you divide it by (x+2) using long division, you'll get -4 as a remainder

Answer 4

honestly, you could have anything like this:
\[\left( x+2 \right) \times p \left( x \right)-4\]
where p(x) is a second degree polynomial

Answer 5

im still not understanding...

Answer 6

let's try one:
\[\left( x+2 \right)^{3}-4 = x ^{3}+2x^{2}+4x+8 -4= x ^{3}+2x^{2}+4x+4\]

Answer 7

i must be annoying, but im still not understanding. what i dont understand is that someone else on here told me that i need to use the formula f(x) = (x+2)(ax^2 + bx + c) - 4 and that i can plug in any number for a b and c besides 0. i followed what they said but im confused as to if i did it right. and now i am seeing what you are saying and its only making me more confused

Answer 8

what they told you was correct, I said the same thing using slightly different language.
do you know we write number using the division algorithm?
n = p*q + r?
ex. for n = 12 and p = 5, we get
12 = 5*2 + 2, where 2 is the remainder.

Now, I ask you to give me a number that when divided by 5 has a remainder of 3. Can you do that?

Answer 9

13?

Answer 10

okay and another?

Answer 11

18?

Answer 12

so, the same thing is going on with the polynomials.

p(x) =d(x)*q(x) + r(x)

this is how it looks for polynomials.

in yur case, they want d(x) = (x+2) and r(x) = -4 but they also want the degree of p(x) to be 3 so q(x) must be a second degree polnomial... and any would do so long as the coefficient on the x squared term isn't 0 (no longer a quadratic).

so you can choose any quadratic (with the restriction previously given) and multiply it by (x+2), subtract 4 and that's a p(x) that'll work.

Answer 13

okay so what about this one...

\[x^3+4x^2+x-10\]

Answer 14

if that doesnt work i think im hopeless

Answer 15

i'm sorry, i never checked your original. let's check this one 1st and then the original.

Answer 16

for your original we only need to divide the quadratic term by (x+2) since we know (x+2) divides itself...

this one doesn't work