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

1. ## 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.

2. ## 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?

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

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

4. ## 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

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

How did they get this?

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

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

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

$5 = -30k$

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

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

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

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

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

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### a polynomial power given 3 with real coefficients

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