Question

The algebraic equation of degree 4 whose roots are the translates of the roots of the equation \( x^4 + 5x^3 + 6x^2 + 7x + 9 = 0 \) by \( -1 \) is:

• A. \( x^4 + 3x^3 - 3x^2 + 6x + 4 = 0 \)
• B. \( x^4 + 9x^3 + 27x^2 + 38x + 28 = 0 \)
• C. \( x^4 + 5x^3 + 6x^2 + 7x + 9 = 0 \)
• D. \( x^4 + 5x^3 + 6x^2 - 7x + 9 = 0 \)

Hint:

To translate the roots of a polynomial by \( -1 \), substitute \( x + 1 \) into the equation and expand the terms.

Answer 1

The Correct Option is B

We are given the equation: \[ x^4 + 5x^3 + 6x^2 + 7x + 9 = 0. \] Let the roots of this equation be \( r_1, r_2, r_3, r_4 \). The equation can then be written as: \[ (x - r_1)(x - r_2)(x - r_3)(x - r_4) = 0. \] We are asked to find the equation whose roots are the translates of these roots by \( -1 \). This means the new roots will be \( r_1 - 1, r_2 - 1, r_3 - 1, r_4 - 1 \). 
Step 1: To translate the roots by \( -1 \), we substitute \( x + 1 \) for \( x \) in the original equation. This gives the new equation: \[ ((x + 1) - r_1)((x + 1) - r_2)((x + 1) - r_3)((x + 1) - r_4) = 0. \] We now expand this expression by substituting \( x + 1 \) into the equation \( x^4 + 5x^3 + 6x^2 + 7x + 9 = 0 \). 
Step 2: Substitute \( x + 1 \) into the original equation: \[ f(x + 1) = (x + 1)^4 + 5(x + 1)^3 + 6(x + 1)^2 + 7(x + 1) + 9. \] Now expand each term: \[ (x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1, \] \[ 5(x + 1)^3 = 5(x^3 + 3x^2 + 3x + 1) = 5x^3 + 15x^2 + 15x + 5, \] \[ 6(x + 1)^2 = 6(x^2 + 2x + 1) = 6x^2 + 12x + 6, \] \[ 7(x + 1) = 7x + 7. \] Thus, the expanded equation is: \[ x^4 + 4x^3 + 6x^2 + 4x + 1 + 5x^3 + 15x^2 + 15x + 5 + 6x^2 + 12x + 6 + 7x + 7 + 9. \] Now combine like terms: \[ x^4 + (4x^3 + 5x^3) + (6x^2 + 15x^2 + 6x^2) + (4x + 15x + 12x + 7x) + (1 + 5 + 6 + 7 + 9). \] This simplifies to: \[ x^4 + 9x^3 + 27x^2 + 38x + 28 = 0. \] Thus, the required equation is \( x^4 + 9x^3 + 27x^2 + 38x + 28 = 0 \).