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.

Respuesta :

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.

Using the Factor Theorem, the polynomial is given by:

[tex]p(x) = x^3 + 13x^2 + 47x + 35[/tex]

The Factor Theorem states that if a polynomial has roots [tex]x_1, x_2, ..., x_n[/tex], it can be written as:

[tex]p(x) = a(x - x_1)(x - x_2)(x - 3)[/tex]

In which a is the leading coefficient.

In this problem:

  • Leading coefficient of 1, thus [tex]a = 1[/tex].
  • Zeros of [tex]x_1 = -1, x_2 = -5, x_3 = -7[/tex].

Then:

[tex]p(x) = (x + 1)(x + 5)(x + 7)[/tex]

[tex]p(x) = (x^2 + 6x + 5)(x + 7)[/tex]

[tex]p(x) = x^3 + 13x^2 + 47x + 35[/tex]

A similar problem is given at https://brainly.com/question/24380382