What are all the roots of x^3+x^2+4x+4=0

Question

Exponents are marked as x^n, n being the exponent

Answer 1

Use the Rational Roots Theorem to obtain possible roots: (factors of the constant)/(factors of the leading coefficient) = (±1, ±2, ±4)/(±1) = ±1, ±2, ±4

Use the Remainder Theorem to find a root: f(x) = 0 is the remainder of a root

f(1) = 1³ + 1² + 4(1) + 4 = 10, x = 1 is not a root

f(-1) = (-1)³ + (-1)² + 4(-1) + 4 = 0, x = -1 is a root

Use synthetic division or long division to find the other roots:

"-1 | 1 1 4 4 " coefficients and constant of the original equation in descending order

" + ↓ -1 0 -4 " drop down the first number (1) and multiply it by -1 = -1 and put it in the 2nd row. Add.

" 1 0 4 0 " multiply the sum = 0 by -1 = 0 and put it in the 2nd row to the right of the last number. Add 4 + 0 = 4. Multiply the sum by -1 = -4 and put it in the 2nd row to the right of the last number. Add. This last number is the remainder = 0. This means that -1 is a root of the original equation.

The last row (the sum) represents the coefficients of the result. It is always one degree less than the original coefficients: 1x² + 0x + 4 with no remainder.

f(x) = (x + 1)(x² + 4)

x² + 4 yields complex roots:

x² + 4 = 0

x² = -4

x = ± sqrt(-4)

x = ± 4i

Factored equation: f(x) = (x + 1)(x - 4i)(x + 4i)

Roots: x = -1, -4i, 4i