# Algebraic Synthetic Division

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In problems 1 through 4, write out the polynomial that has the listed factors:

Example: (x), (x-1) ANS: x2-x

1. (x-2), (x+3)

2. x, (x-2), (x-1)

3. (x-2i), (x+2i)

4. x, (x-1), (x+1), (x-2)

In problems 5 through 8, write the polynominal having the listed roots:

5. i

6. 2, 1, -1

7. 1, 3, 2, -1

8. -2, +3, i

In problems 9 through 11, use synthetic substitution to find f(-3) and f(4).

In problems 12 through 15, you are given a polynomial and one of its factors. Find the remaining factors of the polynomial. Some factors may not be binomials.

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##### Solution Summary

The solution examines algebraic synthetic divisions.

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In problems 1 through 4, write out the polynomial that has the listed factors:

Example: (x),(x-1) ANS: x2-x

1. (x-2),(x+3)

Solution. We assume that the leading coefficient of the polynomial is 1. So, the polynomial is

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

2. x, (x-2), (x-1)

Solution. We assume that the leading coefficient of the polynomial is 1. So, the polynomial is

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

3. (x-2i), (x+2i)

Solution. We assume that the leading coefficient of the polynomial is 1. So, the polynomial is

(x - 2i)(x + 2i) = x^2 + 4

4. x, (x-1), (x+1), (x-2)

Solution. We assume that the leading coefficient of the polynomial is 1. So, the polynomial is

x(x-1)(x + 1)(x - 2) = x^4 - 2x^3 - x^2 + 2x

In problems 5 through 8, write the polynominal having the listed roots:

5. i

Solution. We assume that the leading coefficient of the polynomial is 1. We know that if i is a root, then -i is another root. So, the polynomial is

(x-i)(x+1) = x^2 + 1

6. 2, 1, -1

Solution. We assume that the leading coefficient of the polynomial is 1. So, the polynomial is

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

7. 1, 3, 2, -1

Solution. We assume that the ...

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
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