# Algebraic Synthetic Division

See the attached file.

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 Preview

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 ...

#### Solution Summary

The solution examines algebraic synthetic divisions.