Mathematics, 17.03.2021 23:50 marahkotelman

Build a polynomial of degree 3, with real coefficients, and with zeros i and 1.

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Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

Answer from: Quest

98%

step-by-step explanation:

Answer from: Quest

sasaasasa

step-by-step explanation:

sasasdfubisdbjvkzc

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Build a polynomial of degree 3, with real coefficients, and with zeros i and 1. ...

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