Question

Is
3x+2 a factor of 6x^4-11x^3-7x^2+8x+3


?

A
Yes

B
No

Answer 1

To determine whether \(3x + 2\) is a factor of \(6x^4 - 11x^3 - 7x^2 + 8x + 3\), we can use polynomial long division or synthetic division. However, a quicker method is to check if the polynomial evaluates to zero when the root of \(3x + 2\) is substituted.

First, let's find the root of \(3x + 2 = 0\):

\[
3x + 2 = 0 \implies 3x = -2 \implies x = -\frac{2}{3}
\]

Now we will substitute \(x = -\frac{2}{3}\) into the polynomial \(6x^4 - 11x^3 - 7x^2 + 8x + 3\):

1. Calculate \(6\left(-\frac{2}{3}\right)^4\):
\[
6\left(-\frac{2}{3}\right)^4 = 6 \times \frac{16}{81} = \frac{96}{81}
\]

2. Calculate \(-11\left(-\frac{2}{3}\right)^3\):
\[
-11\left(-\frac{2}{3}\right)^3 = -11 \times -\frac{8}{27} = \frac{88}{27}
\]

3. Calculate \(-7\left(-\frac{2}{3}\right)^2\):
\[
-7\left(-\frac{2}{3}\right)^2 = -7 \times \frac{4}{9} = -\frac{28}{9}
\]

4. Calculate \(8\left(-\frac{2}{3}\right)\):
\[
8\left(-\frac{2}{3}\right) = -\frac{16}{3}
\]

5. The constant term is \(3\) which can be written as \(\frac{243}{81}\) for common denominator purposes.

Summing these values together:
\[
\frac{96}{81} + \frac{88}{27} - \frac{28}{9} - \frac{16}{3} + 3
\]
We convert everything to the common denominator of \(81\):
\[
\frac{96}{81} + \frac{264}{81} - \frac{252}{81} - \frac{432}{81} + \frac{243}{81}
\]

Now summing these:
\[
\frac{96 + 264 - 252 - 432 + 243}{81} = \frac{-81}{81} = -1
\]

Since the result is \(-1\), it is not equal to zero. Therefore, \(3x + 2\) is **not** a factor of the polynomial.

The answer is:

**B. No**