Question

which of the following is a possible zero of the equation: f(x)=2x³ - 3x² - 11x + 6 ○ ±11 ○ ±2/3 ○ ±4 ○ ±3

Explanation:

To find the possible rational zeros of a polynomial \( f(x) = a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0 \), we use the Rational Root Theorem. The theorem states that any possible rational zero, \( \frac{p}{q} \), is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \).

Step 1: Identify \( a_0 \) and \( a_n \)

For the polynomial \( f(x)=2x^3 - 3x^2 - 11x + 6 \), the constant term \( a_0 = 6 \) and the leading coefficient \( a_n=2 \).

Step 2: Find factors of \( a_0 \) and \( a_n \)

The factors of \( a_0 = 6 \) are \( \pm1,\pm2,\pm3,\pm6 \).
The factors of \( a_n = 2 \) are \( \pm1,\pm2 \).

Step 3: Determine possible rational zeros

Using the Rational Root Theorem, the possible rational zeros are \( \frac{p}{q} \), where \( p \) is a factor of \( 6 \) and \( q \) is a factor of \( 2 \). So the possible rational zeros are \( \pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2} \). Wait, but let's check the given options. The options are \( \pm11,\pm\frac{2}{3},\pm4,\pm3 \). Among these, \( \pm3 \) is a factor of \( 6 \) (since when \( q = 1 \), \( \frac{3}{1}=3 \) and \( \frac{- 3}{1}=-3 \)). Let's verify by plugging \( x = 3 \) into \( f(x) \):

\( f(3)=2(3)^3-3(3)^2 - 11(3)+6=2\times27-3\times9-33 + 6=54 - 27-33 + 6=(54 + 6)-(27 + 33)=60 - 60 = 0 \). So \( x = 3 \) is a zero. Also, from the possible rational zeros, \( \pm3 \) is a possible zero (since \( 3 \) is a factor of \( 6 \) and \( 1 \) is a factor of \( 2 \), \( \frac{3}{1}=3 \)). The other options: \( \pm11 \): 11 is not a factor of \( 6 \), so not possible. \( \pm\frac{2}{3} \): \( 2 \) is a factor of \( 6 \)? No, \( q = 3 \) is not a factor of \( 2 \), so \( \frac{2}{3} \) is not a possible zero from Rational Root Theorem. \( \pm4 \): 4 is not a factor of \( 6 \), so not possible. So \( \pm3 \) is a possible zero.

Answer:

\(\pm3\) (i.e., the option with \(\pm3\))