Sure, let's check if [tex]\(7 + 3x\)[/tex] is a factor of [tex]\(3x^3 + 7x\)[/tex]. To do this, we perform polynomial division and see if the remainder is zero.
1. Express the Polynomial: Write the given polynomial [tex]\(3x^3 + 7x\)[/tex].
2. Potential Factor: The potential factor is [tex]\(7 + 3x\)[/tex].
3. Long Division: We will perform polynomial long division to divide [tex]\(3x^3 + 7x\)[/tex] by [tex]\(7 + 3x\)[/tex].
[tex]\[
\frac{3x^3 + 7x}{7 + 3x}
\][/tex]
Start by expressing the polynomial division in the format:
[tex]\[
3x^3 + 7x = (7 + 3x) \cdot Q(x) + R(x)
\][/tex]
where [tex]\(Q(x)\)[/tex] is the quotient and [tex]\(R(x)\)[/tex] is the remainder.
4. Find the Quotient [tex]\(Q(x)\)[/tex]:
- Compare the leading term [tex]\(3x^3\)[/tex] of the dividend with the leading term [tex]\(3x\)[/tex] of the divisor:
The first term of the quotient [tex]\(Q(x)\)[/tex] should be [tex]\(\frac{3x^3}{3x} = x^2\)[/tex].
- Multiply the entire divisor [tex]\(7 + 3x\)[/tex] by this term [tex]\(x^2\)[/tex]:
[tex]\((7 + 3x) \cdot x^2 = 7x^2 + 3x^3\)[/tex].
- Subtract this result from the original polynomial:
[tex]\[
3x^3 + 0x^2 + 7x - (3x^3 + 7x^2) = 0x^3 - 7x^2 + 7x.
\][/tex]
- Compare the new leading term [tex]\(-7x^2\)[/tex] with the leading term [tex]\(3x\)[/tex] of the divisor:
The next term of the quotient [tex]\(Q(x)\)[/tex] should be [tex]\(\frac{-7x^2}{3x} = -\frac{7}{3}x\)[/tex].
- Multiply the entire divisor [tex]\(7 + 3x\)[/tex] by this term [tex]\(-\frac{7}{3}x\)[/tex]:
[tex]\[
(7 + 3x) \cdot -\frac{7}{3}x = -\frac{49}{3}x - 7x^2.
\][/tex]
- Subtract this result from the new polynomial:
[tex]\[
-7x^2 + 7x - (- 7x^2 - \frac{49}{3}x) = 0x^2 + \left(7x + \frac{49}{3}x\right)= 7x + \frac{49}{3}x.
\][/tex]
Combining the terms, we get: [tex]\( \frac{7}{3}x \left(1 + 7\right)\)[/tex].
This term is not zero, thus indicating we have a remainder.
5. Since the remainder is not zero, [tex]\(7 + 3x\)[/tex] is not a factor of [tex]\(3x^3 + 7x\)[/tex].
Therefore, [tex]\(7 + 3x\)[/tex] is not a factor of [tex]\(3x^3 + 7x\)[/tex], and the remainder is [tex]\(x(3x^2 + 7)\)[/tex].
Thus, the final answer is:
[tex]\[
(False, \text{Mod}(x(3x^2 + 7), 3x + 7))
\][/tex]
