Question

write a polynomial function in standard form with zeros at -4,1, and 1

Answer 1

if 'a' is a zero of f(x) then (x-a) is the factor.
hence we have 3 factors here
f(x)=(x+4)(x-1)(x-1)
just multiply them out, and u will get your polynomial :)

Answer 2

any chance you can do this one for me and break down the steps? i knew as far as (x+4)(x-1)(x-1) but after that, i am lost :/

Answer 3

sure :)
(x+4)(x*x - x*1 - x*1 + (-1)*(-1))
= (x+4)(x^2-2x+1)
=(x*x^2 -2x*x +x +4x^2 -8x +4)
f(x)=(x^3 +2x^2 -7x +4)
if u have doubt in any step, u can ask.... :)

Answer 4

ok i am going to enter this in the answer option real quick and if its ok with you,can you stick around for the next one and let me work it out and see if i am right?

Answer 5

sure, go ahead :)

Answer 6

nooo! it didnt give me another one..... make one up lol i will do it on my end

Answer 7

lol! ok.
write a polynomial function in standard form with zeros at 0,-3, and 3.
this is simpler.
if u want little bit difficult, take 2 instead of 0.

Answer 8

sure... i will use the 2 instead

Answer 9

ugh! i got as far as (x-2)(x+3)(x-3)and then
(x-2)(
and then i am lost.. im not sure what order the rest of it goes... obviously i use the distributive prop but which one do i start with and pair it up with what.... i hate math sometimes lol

Answer 10

u can start any one to pair with...
(a+b)(b+c) = ab + ac + bb + bc
or
(a+b)(b+c) = bb + bc+ab + ac
just go slowly.....u will get it. :)

Answer 11

ok im gonna try it again

Answer 12

and post your step, every step, so that i can verify...

Answer 13

im assuming that when we use a,b,c that means that in this equation a=-2 b=3 and c=-3?

Answer 14

am i totally off base with this?

Answer 15

i will write first step, u continue:
(x-2)(x-3)(x+3)
= (x-2) ( x*x +x*3 ..... continue

Answer 16

u got those first two terms ?? any doubts ?

Answer 17

see, this is where i get lost... using all those x's. i see the (x*x +x*3 but what am i using to get it? am i no longer using (x-2) ? am i multiplying the rest of the equation without using x-2

Answer 18

i promise i am not trying to be difficult... i think i am over thinking this and its making me nuts

Answer 19

we will use (x-2) afterwards, we go multiplying 2 brackets at a time.
(..)(..)(..)=(..)(....)=(.....)

Answer 20

=(x-2)(x*x+x*3+x*x+x*-3)?

Answer 21

alrighty i give up lol its all good. thanks for your time. i appreciate it... i got over the part that i needed to learn this for.. i just wanted to try and understand it....