Question

find a polynomial function of minimum degree whose zeros are 0,1,-2

Answer 1

There are three, so minimum degree will be 3.

Each zero results in one factor.

A zero of 1 creates a factor of (x-1)

Go!

Answer 2

how can i solve it by using factor theorem

Answer 3

the factors are
\[(x-r_1)(x-r_2)(x-r_2)\]

Answer 4

where \(r_i\) are the roots (zeros, whatever)

Answer 5

write your polynomial in factored form, then multiply it out
as @tkhunny said, one of the factors is \((x-1)\)

Answer 6

what is the factor form of 0

Answer 7

\(x-0\) which on planet earth is written always as just \(x\)

Answer 8

this one is always confusing
most people get the \((x-1)\) and the \((x+2)\) but get confused by the \(0\) part

Answer 9

then multiply
\[x(x-1)(x+2)\] and you are done

Answer 10

yeah I already get the 1 and -2 but i don't understand the 0

Answer 11

so 0 wont be included anymore, @satellite73 ?

Answer 12

(x-0) is equivalent to x @yuihime

Answer 13

@kc_kennylau , as far as i know, there isn't -0

Answer 14

(x-0)(x-1)(x+2) = x(x-1)(x+2) @yuihime

Answer 15

@kc_kennylau thank you so much

Answer 16

We still don't have a function:

f(x) = ???

Answer 17

@satellite73 thanks for your answer

Answer 18

Well I think that is enough for the answer @tkhunny

Answer 19

No, it's not. Look at the problem statement. "find a polynomial function "

You have not satisfied the requirements if you write only "x(x-1)(x+2)"

Answer 20

someone in a bad mood tonight?
how about
\[f(x)=x(x-1)(x+2)\]?

Answer 21

true, you have to FOIl to find the polynomial

Answer 22

how to you foil three terms?

Answer 23

"FOIL" doesn't mean anything. It was fine as soon as we got the "f(x) ="

Answer 24

Yes! Down with Reynolds Wrap.

Answer 25

so by using remainder theorem, how do you solve \[x(x-1)(x+2)\]

Answer 26

You don't "solve" an expression. You will need an equation.

Answer 27

@tkhunny so how do you do it?

Answer 28

I'm not sure what the question is. Please state the "Remainder Theorem".

Answer 29

Remainder Theorem
if a polynomial P(x) is divided by x-c, where c is a real number, then the remainder is P(c)
@tkhunny

Answer 30

Okay, so now what's the question?

Answer 31

divide \[x ^{3}+x ^{2}+x+1\] by x+1

Answer 32

How's your synthetic division?

Answer 33

well, it can be solve by synthetic as well,

Answer 34

Not "as well". That IS the solution.

If \(p(x) = x^{3}+x^{2}+x+1\)
And we wish to divide by x+1, we get.



Thus, \(\dfrac{p(x)}{x+1} = x^{2} + 1\) with no remainder.