Produce a third degree polynomial that has exactly the roots -3 and 5 with y-intercept 1350.

I understand I could start with (x+3)(x-5)^2, but i don't know how to get the y-intercept 1350 with the same conditions. Please help. thank you.

We don't know that 5 is a double root. So,

y = (x+3)(x-5)(x-k) = <powers of x> + 15k

So, since the y-intercept is where x=0, all the <powers of x> go away and we have 15k = 1350
k = 90

y = (x+3)(x-5)(x-90)

To find a third degree polynomial with the given roots and y-intercept, you can use the fact that the roots of a polynomial are the values of x for which the polynomial equals zero.

Let's start with the factors you have already identified: (x+3)(x-5)^2. This polynomial has the roots -3 and 5, but we need to adjust it to ensure that the y-intercept is 1350.

To find the y-intercept, plug in x=0 into the polynomial equation and set it equal to 1350:

\(f(x) = (x+3)(x-5)^2\)
\(f(0) = (0+3)(0-5)^2 = 15^2 = 225\)

So at x=0, the polynomial value is equal to 225, not 1350. To adjust the y-intercept, we need to introduce a constant factor that will make the polynomial equal to 1350 when x=0.

Let's call this constant factor "a". To find the value of "a", we can divide 1350 by the polynomial value at x=0, which is 225:

\(a = \frac{1350}{225} = 6\)

Now we can include this constant factor, "a", into our initial polynomial:

\(f(x) = a(x+3)(x-5)^2\)

With the value of "a" equal to 6, the adjusted polynomial becomes:

\(f(x) = 6(x+3)(x-5)^2\)

This third degree polynomial has the roots -3 and 5, and its y-intercept is 1350.