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 $x^4 + 9x^3 + 27x^2 + 38x + 28 = 0$.