Question

what is a cubic polynomial with zeros at 4,-1, and 2 ?

Answer 1

the one that looks like
\[(x-4)(x+1)(x-2)\] after you multiply out

Answer 2

can you show me the process

Answer 3

of multiplying?

Answer 4

or how it got the factors?

Answer 5

of how you got the answer

Answer 6

Process of constructing the polynomial is easy:

if we have a polynomial with roots \(r_1, r_2, ... r_n\) (aka zeros) it can be written in factored form as \[a(x-r_1)(x-r_2)...(x-r_n)\]where \(a\) is a constant (usually 1, unless you need to make the polynomial go through a specific point).

Answer 7

If you look at @satellite73's answer, you can see that the product will = 0 if \(x = 4, x=-1, \text{ or } x = 2\)

Answer 8

i didn't get the answer multiplied out, i got it in factored form
if \(r\) is a zero, then one factor is \(x-r\)
your zeros are \(4,-1,2\) so the factors are \((x-4)(x+1)(x-2)\)

Answer 9

what @whpalmer4 said

Answer 10

This is why one approach to finding the zeros (if we don't know them, as we do in this problem) is to try to factor the equation. We set each factor = 0, and solve to find the zeros. If you think about it, you should agree that we should be able to construct a polynomial in this way that will =0 only at the spots where we have zeros, and nowhere else

Answer 11

Does that all make sense? If I haven't confused you yet, I can certainly try again :-)

Answer 12

as for multiplying
\[(x-4)(x+1)(x-2)\] you just have to grind it till you find it
first step would be multiply \((x-4)(x+1)\) and when you get done, multiply the result by \(x-2\)

Answer 13

"grind it till you find it" — ah, a new addition to my collection, right next to "plug'n chug" :-)

Answer 14

im so confussed right now.