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: $(x^2-3x+2)(x^2-7x+12) = x^4-7x^3+12x^2-3x^3+21x^2-36x+2x^2-14x+24$ $= x^4-10x^3+35x^2-50x+24$ Comparing the 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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