Question

What is the value of the leading coefficient a if the polynomial function P(x) = a(x + b)2(x − c) has multiplicity of 2 at the point (–3, 0) and also passes through the points (2, 0) and (0, 36)?

–2
–3
3
36

Answer 1

@Nnesha erm..so like i think your the best at this stuff and i have like 3 others to do after this one think your up to the challenge?

Answer 2

@jim_thompson5910 @jhonyy9 @just_one_last_goodbye

Answer 3

@mathstudent55 think you got this ?

Answer 4

@ShadowLegendX

Answer 5

@zepdrix mind helping?

Answer 6

\[\large\rm f(x)=a(x+b)^2(x-c)\]

Answer 7

x=-3 is one of the zeroes of our function,
which tells us that \(\rm (x+3)\) is a factor of our polynomial.
Since it has multiplicity 2 at this zero, that means \(\rm (x+3)(x+3)=(x+3)^2\) is a factor of our polynomial.

Ok, well we only have one squared term,
so that tells us that \(\rm (x+b)^2=(x+3)^2\)
So your b value?

Answer 8

erm so like i was in the restroom because i got sick so let me catch up to speed real quick @zepdrix

Answer 9

uh -3? yes @zepdrix

Answer 10

@zepdrix finish helping me please.

Answer 11

Hmm it looks like b=+3, yes?

Answer 12

We have another x-intercept at x=2,
So then (x-2) is a factor of our polynomial,
that must correspond to the (x-c)

So we have c=2.

Answer 13

\[\large\rm f(x)=a(x+b)^2(x-c)\]becomes,\[\large\rm f(x)=a(x+3)^2(x-2)\]

Answer 14

If we were to expand out all of the brackets,
err I guess we don't need to do that,
let's just try to figure out the constant term.

Answer 15

Hopefully you remember how to FOIL,\[\large\rm f(x)=a(x^2+6x+9)(x-2)\]So our constant term is the a, times the 9, times the -2,\[\large\rm f(x)=a~bunch~of~x~stuff+a(9)(-2)\]

Answer 16

This function passes through the point (0,36),
Plugging that in gives us,\[\large\rm f(x)=36=0~for~all~the~x~stuff+a(9)(-2)\]\[\large\rm 36=a(9)(-2)\]

Answer 17

Too confusing? :)
Lemme know, I used some weird shortcuts in there hehe

Answer 18

say what now i thought it was -3... its -2??? @zepdrix

Answer 19

(-3,0) tells us that we have an x-intercept of \(\rm x=-3\).
Adding 3 to both sides of this equation gives us \(\rm x+3=0\).
Placing brackets around it just to show that we have a factor of our polynomial.
And we matched it up with the squared term,\[\large\rm (x+\color{orangered}{3})^2\quad=\quad (x+\color{orangered}{b})^2\]I still can't figure out where you're getting -3 from... hmm

Answer 20

You're getting a=-2?
Yah that sounds right :)