Question

Rational roots of the equation \( 2x^4 + x^3 - 11x^2 + x + 2 = 0 \) are:

• A. 1/2, 2
• B. 1/3, 2, -2
• C. 1/2, 2, 3, 4
• D. 1/2, 2, 3, -2

Hint:

The Rational Root Theorem states that if a polynomial equation has a rational solution, it must be of the form \( \frac{p}{q} \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.

Answer 1

The Correct Option is A

Step 1: {Use Rational Root Theorem}
The rational root theorem states that any rational root, say \( \frac{p}{q} \), must be a factor of the constant term (2) divided by a factor of the leading coefficient (2).
Step 2: {Find possible rational roots}
Possible rational roots: \( \pm 1, \pm 2, \pm \frac{1}{2} \). Testing these values, we find that only \( \frac{1}{2} \) and \( 2 \) satisfy the equation.

Answer 2

Step 1: Understand the polynomial
Given equation is:
\[ 2x^4 + x^3 - 11x^2 + x + 2 = 0 \]
We need to find its rational roots.

Step 2: Use Rational Root Theorem
Possible rational roots are factors of constant term (2) divided by factors of leading coefficient (2):
\[ \pm 1, \pm 2, \pm \frac{1}{2} \]

Step 3: Test possible roots
Test \( x = 1 \):
\[ 2(1)^4 + 1 - 11 + 1 + 2 = 2 + 1 - 11 + 1 + 2 = -5 \neq 0 \]
Test \( x = -1 \):
\[ 2(-1)^4 -1 -11 + (-1) + 2 = 2 -1 -11 -1 + 2 = -9 \neq 0 \]
Test \( x = 2 \):
\[ 2(16) + 8 - 44 + 2 + 2 = 32 + 8 - 44 + 2 + 2 = 0 \]
So, \( x = 2 \) is a root.
Test \( x = \frac{1}{2} \):
Calculate:
\[ 2\left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^3 - 11\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 2 = 2 \times \frac{1}{16} + \frac{1}{8} - 11 \times \frac{1}{4} + \frac{1}{2} + 2 \]
\[ = \frac{2}{16} + \frac{1}{8} - \frac{11}{4} + \frac{1}{2} + 2 = \frac{1}{8} + \frac{1}{8} - \frac{11}{4} + \frac{1}{2} + 2 \]
\[ = \frac{2}{8} - \frac{11}{4} + \frac{1}{2} + 2 = \frac{1}{4} - \frac{11}{4} + \frac{1}{2} + 2 = -\frac{10}{4} + \frac{1}{2} + 2 = -\frac{5}{2} + \frac{1}{2} + 2 = -2 + 2 = 0 \]
So, \( x = \frac{1}{2} \) is also a root.

Final Answer: \( \frac{1}{2}, 2 \)