1. ## polynomial

Find a polynomial of degree 3 with zeros at 5, 2, and -2 and which has a leading coefficient of -3.

Find a polynomial of degree 4 that intersects the x-axis ONLY at 2 and -3.

2. 1)The first one is quite simple. Write the roots as factors first:
$\displaystyle f(x) = (x-5)(x-2)(x+2)$

Now the leading coefficient of this polynomial is 1. To make the leading coefficient -3, just multiply -3 to the equation:
$\displaystyle f(x) = -3(x-5)(x-2)(x+2)$

Expand that and you should get your polynomial.

2) Having points of intersection (zeroes) less than the degree of the polynomial indicates that there are some double roots (roots that occur twice)

In your polynomial, you only have two roots (2 and -3). But a polynomial with degree 4 should have 4 roots. We can then conclude that 2 and -3 are both double roots.

Written in factored form it looks like this: $\displaystyle f(x) = (x-2)^2(x+3)^2$

Expand that to get your polynomial.

3. Couldn't the second part also be a polynomial with a triple root? ($\displaystyle (x-2)^3(x+3)$ or $\displaystyle (x-2)(x+3)^3$)

4. Yes. It could also mean that there is a triple root...Thanks for pointing it out.