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Question 16 (1 point) If $6x^4 - 2x^3 - 21x^2 + 7x + 8$ is divided by $3x - 1$ to give a quotient of $2x^3 - 7x$ and a remainder of 8, then which of the following is true? a) $\frac{6x^4 - 2x^3 - 21x^2 + 7x + 8}{3x - 1} = 2x^3 - 7x + \frac{8}{3x - 1}$ b) $6x^4 - 2x^3 - 21x^2 + 7x + 8 = (3x - 1)(2x^3 - 7x) + 8$ c) $x \neq \frac{1}{3}$ d) all of the above

          Question 16 (1 point)
If $6x^4 - 2x^3 - 21x^2 + 7x + 8$ is divided by $3x - 1$ to give a quotient of $2x^3 - 7x$ and
a remainder of 8, then which of the following is true?
a) $\frac{6x^4 - 2x^3 - 21x^2 + 7x + 8}{3x - 1} = 2x^3 - 7x + \frac{8}{3x - 1}$
b) $6x^4 - 2x^3 - 21x^2 + 7x + 8 = (3x - 1)(2x^3 - 7x) + 8$
c) $x \neq \frac{1}{3}$
d) all of the above

Question 16 (1 point)
If 6x^4 - 2x^3 - 21x^2 + 7x + 8 is divided by 3x - 1 to give a quotient of 2x^3 - 7x and
a remainder of 8, then which of the following is true?
a) (6x^4 - 2x^3 - 21x^2 + 7x + 8)/(3x - 1) = 2x^3 - 7x + (8)/(3x - 1)
b) 6x^4 - 2x^3 - 21x^2 + 7x + 8 = (3x - 1)(2x^3 - 7x) + 8
c) x ≠(1)/(3)
d) all of the above

Added by Kevin B.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Melissa Munoz

04/19/2023

Step 1

Step 1: The given information states that 6x^4 - 2x^3 - 21x^2 + 7x + 8 is divided by 3x - 1 to give a quotient of 2x^3 - 7x and a remainder of 8. Show more…

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Question 16 (1 point) If 6x^4 - 2x^3 - 21x^2 + 7x + 8 is divided by 3x – 1 to give a quotient of 2x^3 - 7x and a remainder of 8, then which of the following is true? a) $$\frac{6x^4 - 2x^3 - 21x^2 + 7x + 8}{3x - 1} = 2x^3 - 7x + \frac{8}{3x - 1}$$ b) 6x^4 - 2x^3 - 21x^2 + 7x + 8 = (3x - 1)(2x^3 - 7x) + 8 c) x ≠ 1/3 d) all of the above