# Contructing Polynomial Functions

• June 19th 2008, 03:50 PM
mankvill
Contructing Polynomial Functions
I have to make a polynomial function that uses these:

Third-degree, with zeros of -3, -1, and 2, and passes through the point (4, 7)

(Angry)
• June 19th 2008, 03:52 PM
Moo
Hello !

Quote:

Originally Posted by mankvill
I have to make a polynomial function that uses these:

Third-degree, with zeros of -3, -1, and 2, and passes through the point (4, 7)

(Angry)

If -3, -1 and 2 are zeros of the polynomial P(x), then x-(-3)=x+3, x+1 and x-2 divide P(x).

But (x+3)(x+1)(x-2) is already a polynomial of degree 3. So P(x) is just proportional to (x+3)(x+1)(x-2).

$P(x)=a(x+3)(x+1)(x-2)$

But we know that $P(4)=7$. Substituting, you can get a (Wink)
• June 19th 2008, 03:55 PM
mankvill
...I'm confused. What is the answer, then?

(Doh)
• June 19th 2008, 04:03 PM
Moo
Quote:

Originally Posted by mankvill
...I'm confused. What is the answer, then?

(Doh)

Because (4,7) belongs to the polynomial function, $7=P(4)=a(4+3)(4+1)(4-2)=a*7*5*2=70a$

----> $a=\frac 1{10}$

$P(x)=\frac 1{10}(x+3)(x+1)(x-2)$

Develop to get the complete polynomial
• June 19th 2008, 04:07 PM
mankvill
$x^3/10 + x^2/5 - x/2 - 3/5$

Is this correct?

edit: i don't know how to make a nominator and denominator, obviously
• June 19th 2008, 11:16 PM
Moo
Quote:

Originally Posted by mankvill
$x^3/10 + x^2/5 - x/2 - 3/5$

Is this correct?

Yes (Nod)
And when checking, we get P(-3)=P(-1)=P(2)=0 and P(4)=7, which is exactly what we wanted (Wink)

Quote:

edit: i don't know how to make a nominator and denominator, obviously
\frac{numerator}{denominator} :)