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2 years ago

Write a polynomial function with zeros -3,5,8

Mathematics

1 answer:

klemol [59]2 years ago

3 0

Answer:

x^3-10x^2+x+120

Step-by-step explanation:

Assuming you mean roots -3, 5, 8

These happen when we have (x+3)(x-5)(x-8)=0

Expand this

(x^2-2x-15)(x-8)=0

=x^3-2x^2-15x-8x^2+16x+120

=x^3-10x^2+x+120

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Lolita deposited 500 in her savings account that earns 5% intrest compounded anually. She forgot about it until now, 15 years la

Sonja [21]

Answer:

P = 1039.5

Step-by-step explanation:

Given:-

- The initial amount deposited, Po = 500

- The interest rate applied, I = 5% compounded annually

Find:-

- The amount on her bank statement after 15 years?

Solution:-

- We see that the principal amount increases every year and no transactions have been made in the course of 15 years.

The total amount left in her savings account would be given by the following formula:

P = Po * ( 1 + I/100 )^n

- Where, n = number of years passed since deposit. (15 years)

P = 500 * ( 1 + 5/100 )^15

P = 500 * (1.05)^15

P = 1039.5

6 0

3 years ago

2^4n × 2^ 2n = 512<br> What is the value of n.

Marina86 [1]

Step-by-step explanation:

${2}^{4n} \times {2}^{2n} = 512$

${2}^{4n} \times {2}^{2n} = {2}^{9}$

${2}^{4n + 2n} = {2}^{9}$

$6n = 9$

$n = \frac{3}{2}$

8 0

2 years ago

Read 2 more answers

What is the area of a field that is 132.4m long by 28m wide?

svetlana [45]

Answer:

Step-by-step explanation:

its 132.4 times 28

=3707.2

7 0

3 years ago

What is vertex of x^2+5x-3

Art [367]

$\text{The vertex of}\ ax^2+bx+c:\\\\(h,\ k),\ \text{where}\ h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(-b^2-4ac)}{4a}$

$\text{We have:}\\f(x)=x^2+5x-3\\\\a=1,\ b=5,\ c=-3\\\\h=\dfrac{-5}{2\cdot1}=\dfrac{-5}{2}=-2.5\\\\k=f(-2.5)=(-2.5)^2+5(-2.5)-3=6.25-12.5-3=-9.25$

$\text{Answer: the vertex is in}\ (-2.5;\ -9.25)$

6 0

3 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago

Write a polynomial function with zeros -3,5,8​

Write a polynomial function with zeros -3,5,8