Question

HELP. Find a cubic function with the given zeros.

Answer 1

\[\sqrt{5}, -\sqrt{5}, -7\]

Answer 2

Those are the zeros^^. My choices are:
f(x) = x^3 - 7x^2 - 5x - 35
f(x) = x^3 + 7x^2 - 5x + 35
f(x) = x^3 + 7x^2 - 5x - 35
f(x) = x^3 + 7x^2 + 5x - 35

Answer 3

expand:\[(x^2 - 5)(x + 7) = 0\]

see how plugging in either of those zeros will make either bracket 0

Answer 4

another way of writing it (which i probably should have chosen is:\[(x + \sqrt5)(x-\sqrt5)(x+7)\]

Answer 5

OH! Oh my gosh.. You made it seem so simple. Wow. I feel ridiculous. Because those are opposites, and... wow.

Answer 6

hehe. glad i could help ^_^

Answer 7

Wait, how do I know which answer choice that would be. Do I have to work them all out, or do you have a made me feel dumb answer for that, too? :p

Answer 8

I can just multiply that out vvv, right?\[ (x^{2}-\sqrt{5}) \times (x^{2}+\sqrt{5}) \times (x+7)\] @Euler271

Answer 9

Oops, not x^2..

Answer 10

yes you can. multiplying it should give you the polynomial you're looking for

Answer 11

you expand the brackets and your result will be one of the choices

Answer 12

Okay, I got x^3+7x^2-5x-35. Sound about right? :D

Answer 13

yes that is the right one

Answer 14

Yay. I have another one. But it's not like this one. Think you're up for it?

Answer 15

ya. bring it on

Answer 16

Find the zeros of the polynomial function and state the multiplicity of each.

f(x) = 4(x + 7)^2(x - 7)^3

Answer 17

Choices:
A) 4, multiplicity 1; -7, multiplicity 3; 7, multiplicity 3
B) -7, multiplicity 3; 7, multiplicity 2
C) 4, multiplicity 1; 7, multiplicity 1; -7, multiplicity 1
D) -7, multiplicity 2; 7, multiplicity 3

Answer 18

which do you think? this can be re-written as:\[f(x) = 4(x+7)(x+7)(x-7)(x-7)(x-7)\]

Answer 19

Oh. Gosh, you made this one easy, too. You need to move in with me, bro. So that would make it... The four doesnt count, right? So D?

Answer 20

lol ^_^
that's right: 4 is just a constant

D is correct

Answer 21

Yay! I feel smart. It's really hard to do all this on FLVS. -_- Thanks! I'll probably be back. This girl's got lotsa tests to take.

Answer 22

glad i could help ^_^

Answer 23

You still there? I have to do (x-5)^5. Would it make sense to break this up into (x-5)^2, and (x-5)^3, and then multiply those, or is there an easier way?

Answer 24

@Euler271

Answer 25

sorry for delay.

you can, but there is an easier way: using pascal's triangle. every heard of it?

http://i.stack.imgur.com/nsj3h.gif

Answer 26

the numbers are the coefficients of the polynomial.

you must use the row with "5" in it since we have (a-b)^5

my explanation will be sloppy because it's typed but:

\[(a-b)^5 = 1a^5b^0 - 5a^4b^1 + 10a^3b^2 - 10a^2b^3 + 5a^1b^4 - 1a^0b^5\]
notice that the coefficients are pascal's triangle numbers and that the power of "a" decreases every term by 1 and the power of "b" increases by 1

if we had (a+b)^5, ALL terms would be positive. having (a-b)^5, the minus sign makes every term's sign alternate from + to -.

do you want to try it or should i do it?

replace a by x and b by 5

Answer 27

I tried it and got... \[x^{5}-25x^{4}+250x^{3}-1250x^{2}+3125x-3125\]

Answer 28

that is the correct answer ^_^

good site for double checking: http://www.wolframalpha.com/input/?i=%28x-5%29%5E5