Question

perform the operation:
\\((12x^4 + 3x^2 + 6x + 1) \div (3x)\\)
a. \\(4x^3 + x + 2 + \frac{1}{3}x\\)
b. \\(4x^3 + 3x^2 + 6x + 1\\)
c. \\(4x^3 + x + 2 + \frac{1}{3x}\\)
d. \\(4x^4 + x^2 + 2x + \frac{1}{3x}\\)
\\(\bigcirc\\) a
\\(\bigcirc\\) b
\\(\bigcirc\\) c
\\(\bigcirc\\) d

Explanation:

Step1: Divide each term

We divide each term in the polynomial \(12x^{4}+3x^{2}+6x + 1\) by \(3x\).

• For the first term: \(\frac{12x^{4}}{3x}=\frac{12}{3}\times\frac{x^{4}}{x}=4x^{3}\) (using the rule \(\frac{x^{m}}{x^{n}}=x^{m - n}\), here \(m = 4\), \(n = 1\), so \(x^{4-1}=x^{3}\))
• For the second term: \(\frac{3x^{2}}{3x}=\frac{3}{3}\times\frac{x^{2}}{x}=x\) (using the same rule, \(m = 2\), \(n = 1\), so \(x^{2 - 1}=x\))
• For the third term: \(\frac{6x}{3x}=\frac{6}{3}\times\frac{x}{x}=2\) (since \(\frac{x}{x}=1\))
• For the fourth term: \(\frac{1}{3x}\) (because \(1\) divided by \(3x\) is \(\frac{1}{3x}\))

Step2: Combine the results

Combining all the divided terms, we get \(4x^{3}+x + 2+\frac{1}{3x}\), which matches option C.

Answer:

C. \( 4x^{3}+x + 2+\frac{1}{3x} \)