Question

Use synthetic division and the remainder theorem to find P(x).
P(x)=x^3+7x^2+4x; x= -2

Answer 1

Do you know what the remainder theorem implies?

Answer 2

I read some on a lesson about it... does it include the long division?

Answer 3

No, the long division is unnecessary. It means that the remainder that you get from synthetic division is the same as evaluating the function at that value of x. This is why synthetic division is also called synthetic substitution.

Answer 4

It, together with the factor-root theorem, is a key tool to solving higher-order polynomials.

Answer 5

For this function, you can factor out an x, then do synthetic division on the quadratic that is left. The remainder at the end = P(2).

Answer 6

so i would divide the function by 2 using synthetic division to get the answer?

Answer 7

yep.

Answer 8

You can check your solution by subbing x=2 into the function to make sure you get the same thing.

Answer 9

okay so i divided it... i got 15 remainder -6??

Answer 10

hmmm, no..

Answer 11

Oh.. hold up, is that x=-2?

Answer 12

*adjusts screen resolution*

Answer 13

but it is a negative 2 haha yea... nowi got the rmainder as 12?

Answer 14

yes, the quotient is x^2 +5x -6 with a remainder of 12, so P(-2)=12.

Answer 15

Okay wonderful!!!! THANK YOU!!

Answer 16

Pop quiz: what would it mean if the remainder was zero?

Answer 17

then x=-2 right??

Answer 18

No, x=-2 is given.
If the remainder of synthetically dividing by -2 was zero, then P(-2)=0 and that means that x=-2 is a root of the polynomial, i.e. it is a solution of the equation P(x)=0.

Answer 19

Ohh i see sorry i didnt get that one this time :)

Answer 20

It'll all come together soon.
When it comes to solving higher-order polynomials, you'll need a combination of:
The fundamental theorem of calculus
The factor-root theorem
The remainder theorem
The rational root theorem
Descartes' rule of signs
and graphing skills.
Practice all of the above and you should be able to solve many polynomial functions (as long as they don't have mostly irrational or complex roots).

Answer 21

Haha great!! Thank you i will write that down :)