Write a polynomial in standard form axn+bxn-1+ ... given the following requirements.

Degree: 3, Zeros at (-1,0), (-5,0) and (-7,0) and y-intercept at (0,35). Leading coefficient is 1.

We are given zeros : (-1,0), (-5,0) and (-7,0).

We can write those zeros as

x=-1, x=-5 and x=-7.

So, the factors of the polynomial would be

(x+1), (x+5) and (x+7).

We also given leading coefficent = 1.

Let us multiply all those factors of the polynomila we got and then multiply by 1 finally.

(x+1) * (x+5) * (x+7).

Let us foil first two factors (x+1) * (x+5) first, we get

x^2 + 5x + 1x + 5 = x^2 + 6x + 5.

Therefore, (x+1) * (x+5) * (x+7) = (x^2 + 6x + 5) (x+7).

Let us multiply (x^2 + 6x + 5) and (x+7), we get

(x^2 + 6x + 5) * (x+7) = x^3 + 7x^2 + 6x^2 + 42x + 5x + 35.

Combining like terms 7x^2 + 6x^2 = 13x^2 and 42x + 5x = 47x.

Therefore,

x^3 + 7x^2 + 6x^2 + 42x + 5x + 35

= x^3 + 13x^2 + 47x + 35.

If we multiply it by 1, we get same terms.

And also if we plug x=0, we get

(0) ^3 + 13 (0) ^2 + 47 (0) + 35 = 0+0+0+35 = 35.

We can see that y-intercept is 35 there, because we get 35 on plugging x=0.

Therefore, x^3 + 13x^2 + 47x + 35 is the polynomial in standard form that fulfil all the given requirements.