Question

How to solve for 'a' in this polynomial?
8= a(x+2)^4(x-3)^3
I missed 2 days in school because I was sick :/

Answer 1

ignore most of it, and try to visualize what the constant will be when you multiply all this junk out

Answer 2

Do I have to go (x+2)(x+2)(x+2)(x+2) and so on? or is there an easier way?

Answer 3

\[8= a(x+2)^4(x-3)^3 \]
\[8=\text{bunch of junk}+a\times 2^4\times (-3)^3\]

Answer 4

Why not divide both sides by \((x+2)^4(x-3)^3\)

Answer 5

oh wait, i am totally wrong
how in the world is that stuff equal to 8?

Answer 6

Then, see what \(x\) has to be at the roots \(3\) and \(-2\) to fill in any gaps.

Answer 7

ignore my answer, go with @wio

Answer 8

Seems like at the roots \(a\) would be undefined.

Answer 9

oh the original was suppose to be f(x)= (x+2)^4 (x-3)^3 - but the question was to find the point (-1, 8), so I figured i would just substitute 8

Answer 10

oooooh!!!

Answer 11

The point is either going to be on the function or not...

Answer 12

ok then you are TOLD \(x=-1\) so the equation is
\[8=a(-1+2)^2(-1-4)^3\]

Answer 13

ok confused again
is there an \(a\) in the problem, or not?

Answer 14

lol uhmmm I'll write the whole equation :P, I was just doing it liek in my notes XD. "Write the equation for a polynomial where a zero at -2 has a multiplicity of 4 and a zero at +3 has a multiplicity of 3 and the graph passes through the point (-1, 8)"

Answer 15

if \[f(x)= (x+2)^4 (x-3)^3 \]
\[f(-1)=(-1+2)^4(-1-4)^3=-5^3=-125\] which is not \(8\)

Answer 16

ok back to square one
you were right to have the \(a\) there

Answer 17

\[f(x)= a(x+2)^4 (x-3)^3\] is correct

Answer 18

oh ok :)

Answer 19

then set \[f(-1)=a(-1+2)^4(-1-3)^4=-64a=8\]

Answer 20

solve for \(a\) and you are done

Answer 21

clear more or less?

Answer 22

ohh right I forgot you could substiture the negative 1 for X, now uhmmm how did you get the -64a?

Answer 23

oh nvm

Answer 24

got it

Answer 25

ok thanks :P

Answer 26

yw