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.

Using the Factor Theorem, the polynomial is given by:

The Factor Theorem states that if a polynomial has roots , it can be written as:

In which a is the leading coefficient.

In this problem:

Then:

A similar problem is given at link

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.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)