Find the remainder when x^4-11x^3+44x^2-76x+4 | vedclass

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(A) Let $P(x) = x^4-11x^3+44x^2-76x+48$ and $D(x) = x^2-7x+12$.First,factorize the divisor: $D(x) = (x-3)(x-4)$.By the division algorithm,$P(x) = D(x)

Vedclass · Accepted Answer

$0$

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(A) Let $P(x) = x^4-11x^3+44x^2-76x+48$ and $D(x) = x^2-7x+12$.First,factorize the divisor: $D(x) = (x-3)(x-4)$.By the division algorithm,$P(x) = D(x)Q(x) + R(x)$,where $R(x) = ax+b$ is the remainder.Thus,$P(x) = (x-3)(x-4)Q(x) + ax+b$.For $x=3$: $P(3) = 3^4 - 11(3^3) + 44(3^2) - 76(3) + 48 = 81 - 297 + 396 - 228 + 48 = 0$.So,$3a+b = 0$.For $x=4$: $P(4) = 4^4 - 11(4^3) + 44(4^2) - 76(4) + 48 = 256 - 704 + 704 - 304 + 48 = 0$.So,$4a+b = 0$.Solving the system $3a+b=0$ and $4a+b=0$ gives $a=0$ and $b=0$.Therefore,the remainder is $0$.

Find the remainder when $x^4-11x^3+44x^2-76x+48$ is divided by $x^2-7x+12$.

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