AnswersFL-1200330-Algebra 2 CRSynthetic Division and the Remainder Theorem

Synthetic Division and the Remainder Theorem Answers

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Dividend:

A
2x3 + 11x2 + 18x + 9
B
-6x3 - 15x - 9
C
2x2 + 5x + 3
3

Quotient:

A
2x4 + 11x3 + 18x + 9
B
2x3 + 5x2 + 3x
C
2x2 + 5x + 3
6

Select all of the factors of x3 + 5x2 + 2x – 8.

A
x + 5
B
x – 3
C
x + 4
D
x – 1
E
x + 3
F
x + 2
7

Select all of the roots of x3 + 2x2 – 16x– 32.

A
–1
B
–2
C
–4
D
1
E
4
F
5
13

Select all of the factors of x3 + 5x2 + 2x – 8.

A
x + 5
B
x – 3
C
x + 4
D
x – 1
E
x + 3
F
x + 2
14

One root of f(x) = 2x3 + 9x2 + 7x – 6 is –3. Explain how to find the factors of the polynomial.

Answer:

Since -3 is a root, x + 3 is a factor. Use synthetic division with -3 and the coefficients 2, 9, 7, and -6. The quotient is 2x^2 + 3x - 2, with remainder 0. Factoring the quotient gives (x + 2)(2x - 1), so the factors of the polynomial are (x + 3)(x + 2)(2x - 1).

15

Which of the following facts did you include in your explanation? Select all that apply.

A
Identify the factor x + 3 from the given root –3.
B
Use synthetic division to divide the polynomial by x + 3.
C
Use the bottom row of the synthetic division as coefficients in the quadratic 2x2 + 3x – 2.
D
Factor the quotient to find the two other factors: x + 2 and 2x – 1.
18

A student uses the division shown to divide –3x4 + 15x3 – x + 5 by x – 5, and concludes that x – 5 is not a factor of the polynomial. Describe two errors the student made. Is x – 5 a factor? Explain.

Question illustration
Answer:

The student made two main errors in the synthetic division setup. First, they used -5 instead of 5 as the value for k (the divisor should be 5 since the factor is x - 5). Second, they did not use a placeholder zero for the missing x^2 term in the polynomial. If the division is performed correctly, the remainder is 0, which means x - 5 is indeed a factor of the polynomial.

20

Dividend:

A
2x3 + 11x2 + 18x + 9
B
-6x3 - 15x - 9
C
2x2 + 5x + 3

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