The polynomial equation of degree 5 whose roo | vedclass

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(C) Let the roots of the given equation $x^5-3x^4-x^3+11x^2-12x+4=0$ be $\alpha_i$. We want the equation whose roots are $\beta_i = \alpha_i + 2$. Thi

Vedclass · Accepted Answer

$x^5-13x^4+63x^3-135x^2+108x=0$

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(C) Let the roots of the given equation $x^5-3x^4-x^3+11x^2-12x+4=0$ be $\alpha_i$. We want the equation whose roots are $\beta_i = \alpha_i + 2$. This implies $\alpha_i = \beta_i - 2$. Substituting $x = y - 2$ into the original equation,we get $(y-2)^5 - 3(y-2)^4 - (y-2)^3 + 11(y-2)^2 - 12(y-2) + 4 = 0$. Expanding each term: $(y-2)^5 = y^5 - 10y^4 + 40y^3 - 80y^2 + 80y - 32$ $-3(y-2)^4 = -3(y^4 - 8y^3 + 24y^2 - 32y + 16) = -3y^4 + 24y^3 - 72y^2 + 96y - 48$ $-(y-2)^3 = -(y^3 - 6y^2 + 12y - 8) = -y^3 + 6y^2 - 12y + 8$ $11(y-2)^2 = 11(y^2 - 4y + 4) = 11y^2 - 44y + 44$ $-12(y-2) = -12y + 24$ $+4 = 4$ Summing these: $y^5 + (-10-3)y^4 + (40+24-1)y^3 + (-80-72+6+11)y^2 + (80+96-12-44-12)y + (-32-48+8+44+24+4) = 0$ $y^5 - 13y^4 + 63y^3 - 135y^2 + 108y = 0$. Thus,the equation is $x^5 - 13x^4 + 63x^3 - 135x^2 + 108x = 0$.

The polynomial equation of degree $5$ whose roots are the roots of the equation $x^5-3x^4-x^3+11x^2-12x+4=0$ each increased by $2$,is

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