Find the rational roots of x^4 + 3x^3 + 3x^2 - 3x - 4 = 0

0, 1

1, 2

1, - 1

-1, 2

Answer: - 1, 1

Step-by-step explanation:

To solve the polynomial equation x^4 + 3x^3 + 3x^2 - 3x - 4 = 0, do a quick test of polynomials using 1 and - 1 in place of x.

This gives the sum of coefficients as zero

Testing with 1: 1 + 3 + 3 - 3 - 4 = 0

Testing with - 1: 1 + (-3) + 3 - (-3) - 4

= 1 - 3 + 3 + 3 - 4 = 0

This gives that both (x + 1) and (x - 1) are roots.

(x + 1) (x - 1) = x² - 1 {difference of two squares}

Dividing the polynomial x^4 + 3x^3 + 3x^2 - 3x - 4 by (x² - 1) results in x² - 3x + 4 as quotient. Factorizing this result would give complex roots.

Therefore, x² - 1 = 0

x² = 1

Taking square of both sides of the equation gives

x = ± 1

1 and - 1 are the rational roots to the polynomial.

C. 1,-1

Step-by-step explanation:

The sum of the coefficient is 0 so x=1 is a root and (x-1) as a factor

so x^4+3x^3+3x^2-3X-4 = (x-1) (x^3+4x^2+7x+4)

and the root of that is x=-1 and the factor is (x+1)

SO the answer is: x=1 and x=-1