# Find a polynomial of degree 3 with real coefficients that satisfies...

• Mar 4th 2012, 11:54 AM
SammyAbby
Find a polynomial of degree 3 with real coefficients that satisfies...
Find a polynomial of degree 3 with real coefficients that satisfies the given conditions.?

Zeros of -3,-1, 4, and P(2) = 5

I am attempting this problem in order to study for an upcoming test. I am stuck on this one, and would really like to learn if someone could please assist me, I would very much appreciate it.
• Mar 4th 2012, 01:04 PM
SammyAbby
Re: Find a polynomial of degree 3 with real coefficients that satisfies...
I have multiplied (x + 3)(x+1)(x-4)
to get x^3 - 13x -12
Where do I go from here?
• Mar 4th 2012, 01:36 PM
skeeter
Re: Find a polynomial of degree 3 with real coefficients that satisfies...
Quote:

Originally Posted by SammyAbby
I have multiplied (x + 3)(x+1)(x-4)
to get x^3 - 13x -12
Where do I go from here?

P(x) = k(x+3)(x+1)(x-4)

5 = k(2+3)(2+1)(2-5)

solve for k
• Mar 5th 2012, 11:16 AM
SammyAbby
Re: Find a polynomial of degree 3 with real coefficients that satisfies...
Thank you for your assistance...I am still a bit confused...my book says that the answer is P(x) = (-x^3/6) + (13x/6) + 2
• Mar 5th 2012, 11:17 AM
SammyAbby
Re: Find a polynomial of degree 3 with real coefficients that satisfies...
How did they get this?
• Mar 5th 2012, 03:27 PM
skeeter
Re: Find a polynomial of degree 3 with real coefficients that satisfies...
$\displaystyle P(x) = k(x+3)(x+1)(x-4)$

$\displaystyle 5 = k(2+3)(2+1)(2-4)$

$\displaystyle 5 = -30k$

$\displaystyle k = -\frac{1}{6}$

$\displaystyle P(x) = k(x+3)(x+1)(x-4)$

$\displaystyle P(x) = -\frac{1}{6}(x+3)(x+1)(x-4)$

$\displaystyle P(x) = -\frac{1}{6}(x^3-13x-12)$

$\displaystyle P(x) = -\frac{x^3}{6} + \frac{13x}{6} + 2$