The algebraic equation of degree 4 whose root | vedclass

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(B) To find the equation whose roots are shifted by $-1$,we replace $x$ with $(x - (-1))$,which is $(x+1)$.Substituting $(x+1)$ for $x$ in the origina

Vedclass · Accepted Answer

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

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(B) To find the equation whose roots are shifted by $-1$,we replace $x$ with $(x - (-1))$,which is $(x+1)$.Substituting $(x+1)$ for $x$ in the original equation $f(x) = x^4+5x^3+6x^2+7x+9=0$:$f(x+1) = (x+1)^4 + 5(x+1)^3 + 6(x+1)^2 + 7(x+1) + 9 = 0$Expanding 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$$7(x+1) = 7x + 7$Adding these together:$x^4 + (4+5)x^3 + (6+15+6)x^2 + (4+15+12+7)x + (1+5+6+7+9) = 0$$x^4 + 9x^3 + 27x^2 + 38x + 28 = 0$

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

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