If 1, 2, 3 and 4 are the roots of the equatio | vedclass

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(C) Given that $1, 2, 3, 4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$. Therefore,we can write the polynomial as: $(x-1)(x-2)(x-3)(x-4) = x^

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$10$

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(C) Given that $1, 2, 3, 4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$. Therefore,we can write the polynomial as: $(x-1)(x-2)(x-3)(x-4) = x^4+ax^3+bx^2+cx+d$ Expanding the left side: $((x-1)(x-4))((x-2)(x-3)) = (x^2-5x+4)(x^2-5x+6)$ Let $y = x^2-5x$. Then the expression becomes $(y+4)(y+6) = y^2+10y+24$. Substituting back $y = x^2-5x$: $(x^2-5x)^2 + 10(x^2-5x) + 24 = x^4 - 10x^3 + 25x^2 + 10x^2 - 50x + 24$ $= x^4 - 10x^3 + 35x^2 - 50x + 24$. Comparing coefficients with $x^4+ax^3+bx^2+cx+d=0$,we get: $a = -10, b = 35, c = -50, d = 24$. Now,calculate $a+2b+c$: $a+2b+c = -10 + 2(35) - 50 = -10 + 70 - 50 = 10$.

If $1, 2, 3$ and $4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$,then $a+2b+c$ is equal to

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