Expert: Chanda Walker - 3/27/2009

Question
Hey,

Finally got around to finishing the course and got my yearend Review. I finished the handout for the most part but got stuck on a few questions and was hoping you could explain them.

When the polynomial x^3-kx^2+17x+6 is divided by x-3, the remainder is 12. What is the value of K?

How would I tackle this? I know that X-3=0. X=3.

Do I plug the 3 in? What do I do with the remainder of 12? I've always had the variables.

Determine the equation of each function in factored form, given the following conditions.

A quartic function tangent to the x-axis at (-1,0) and (2,0) which passes through (3,-16).

Is a quartic function a variable to the power of 4? And when they mention tangent do they mean (-1,0)(-1,0) and (2,0)(2,0)

http://www.mathhelpforum.com/math-help/attachments/pre-calculus/6224d1210129085-...

Like what's happening in the 1st quadrant right? What happens then?

For each pair of functions, find f (g(3)), f (g(x)), g (f(3)), g(f(x))

a. f(x)=5-3x, g(x)=4x+1

For g(3) I just plug in 3 wherever I see X I understand that. But what do I do with f(g(x))?

Thanks in advance!

Answer
I've never worked a problem like this so I'm not sure the best approach but I would attack it this way.  If the remainder is 12 then

x^3 + kx^2 + 17x + 6 -12 = 0

x^3 - kx^2 + 17x - 6 = 0

so you'll have a solution that looks like this:

(x - 3)(ax^2 + bx + c) = 0


(x - 3)(ax^2 + bx + c) = x^3 - kx^2 + 17x - 6

now let's multiply that out and solve for b.

ax^3 - 3ax^2 + bx^2 - 3bx +cx - 3c = x^3 - kx^2 + 17x - 6

ax^3 + (b-3a)x^2 + (-3b + c)x - 3c = x^3 - kx^2 + 17x - 6

We can see:
a = 1
c = 2

so
(b - 3)x^2 +(-3b + 2)x = kx^2 + 17x

(-3b + 2) = 17
-3b = 15
b = -5


-k = (b - 3)
-k = (-5 - 3)
k = 8

Hey that wasn't so bad. ;)
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Determine the equation of each function in factored form, given the following conditions.

A quartic function tangent to the x-axis at (-1,0) and (2,0) which passes through (3,-16).

This is a quartic function:
y(x) = ax^4 +bx^3 + cx^2 + dx + e = 0

If it is tangent to the x-axis then we know:
y'(-1) = 0
y'(2) = 0

y'(x) = 4a x^3 + 3b x^2 + 2c x + d
y'(-1) = -4a + 3b - 2c + d = 0   (eq 1)
y'(2) = 32a + 12b + 4c + d = 0   (eq 2)

Since they both equal zero, let's set them equal to each other.  I'm going to skip a few steps here . . .

36a + 9b + 6c = 0


We also know
y(-1) = 0
a - b + c - d + e = 0             (eq 3)

y(2) = 0
16a + 8b + 4c + 2d + 3 = 0        (eq 4)

y(3) = -16
81y + 27b + 9c + 3d + e = 16      (eq 5)

That gives us 5 equations and 5 unknowns.  From here I think it is simply a linear algebra problem.  If you need further help, let me know but I have to attend to other priorities right now.

I don't understand what you are trying to represent with your graph.  I'll check back as soon as I can.

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For each pair of functions, find f (g(3)), f (g(x)), g (f(3)), g(f(x))

a. f(x)=5-3x, g(x)=4x+1

For g(3) I just plug in 3 wherever I see X I understand that. But what do I do with f(g(x))?

g(3) = 13
f(g(3)) = f(13) = 5 - 39 = -34


f(g(x)) = 5 - 3(4x + 1) = 5 - 12x - 3 = 2 - 12x

f(3) = -4
g(f(3)) = g(-4) = -15


g(f(x)) = 4(5 - 3x) + 1 = 20 - 12x + 1 = 21 - 12x

Sorry to run out on that second question, I'll be back though!