If f(x) = x^4 - 2x^3 + 3x^2 - ax + b is divid | vedclass

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(C) Given $f(x) = x^4 - 2x^3 + 3x^2 - ax + b$. By the Remainder Theorem,$f(1) = 5$ and $f(-1) = 19$. For $f(1) = 5$: $1 - 2 + 3 - a + b = 5 \implies 2

Vedclass · Accepted Answer

$10$

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(C) Given $f(x) = x^4 - 2x^3 + 3x^2 - ax + b$. By the Remainder Theorem,$f(1) = 5$ and $f(-1) = 19$. For $f(1) = 5$: $1 - 2 + 3 - a + b = 5 \implies 2 - a + b = 5 \implies b - a = 3$ (Eq. $i$). For $f(-1) = 19$: $1 + 2 + 3 + a + b = 19 \implies 6 + a + b = 19 \implies a + b = 13$ (Eq. $ii$). Adding Eq. $i$ and Eq. $ii$: $(b - a) + (a + b) = 3 + 13 \implies 2b = 16 \implies b = 8$. Substituting $b = 8$ into Eq. $ii$: $a + 8 = 13 \implies a = 5$. Thus,$f(x) = x^4 - 2x^3 + 3x^2 - 5x + 8$. To find the remainder when $f(x)$ is divided by $(x - 2)$,we calculate $f(2)$: $f(2) = (2)^4 - 2(2)^3 + 3(2)^2 - 5(2) + 8$ $f(2) = 16 - 16 + 12 - 10 + 8 = 10$. Therefore,the remainder is $10$.

If $f(x) = x^4 - 2x^3 + 3x^2 - ax + b$ is divided by $(x - 1)$ and $(x + 1)$,the remainders are $5$ and $19$,respectively. If $f(x)$ is divided by $(x - 2)$,the remainder is:

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