Question

The sum of the real roots of the equation \[ x^4-2x^3+x-380=0 \] is

• A. \(-1\)
• B. \(0\)
• C. \(1\)
• D. \(2\)

Hint:

To find real roots of higher degree polynomials, first try rational roots using the factor theorem. Then factorize completely and use the discriminant to identify whether quadratic factors have real roots.

Answer 1

The Correct Option is C

Step 1: Write the polynomial equation.
We need to find the sum of real roots of $x^4 - 2x^3 + x - 380 = 0$.
Step 2: Try to find rational roots using the rational root theorem.
Possible rational roots are factors of 380: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 19, \pm 20, \ldots$ Test $x=5$: $625 - 250 + 5 - 380 = 0$ ✓. So $(x-5)$ is a factor.
Step 3: Divide out $(x-5)$ by synthetic/polynomial division.
$x^4 - 2x^3 + x - 380 = (x-5)(x^3 + 3x^2 + 15x + 76)$. Verify: $(x-5)(x^3+3x^2+15x+76) = x^4+3x^3+15x^2+76x-5x^3-15x^2-75x-380 = x^4-2x^3+x-380$ ✓.
Step 4: Find a root of the cubic $x^3+3x^2+15x+76$.
Test $x = -4$: $-64 + 48 - 60 + 76 = 0$ ✓. So $(x+4)$ is a factor.
Step 5: Divide out $(x+4)$ from the cubic.
$x^3+3x^2+15x+76 = (x+4)(x^2-x+19)$. The discriminant of $x^2-x+19$ is $1-76 = -75 < 0$, so it has no real roots.
Step 6: Sum only the real roots.
The only real roots are $x=5$ and $x=-4$. Their sum is $5+(-4) = 1$. \[ \boxed{1} \]