If the polynomial p(x) = x^{3} - 3x^{2} + x + | vedclass

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(A) According to the division algorithm for polynomials, $p(x) = s(x) \cdot q(x) + r(x)$.Given $p(x) = x^{3} - 3x^{2} + x + 2$, $q(x) = x - 2$, and $r

Vedclass · Accepted Answer

$x^{2} - x + 1$

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(A) According to the division algorithm for polynomials, $p(x) = s(x) \cdot q(x) + r(x)$.Given $p(x) = x^{3} - 3x^{2} + x + 2$, $q(x) = x - 2$, and $r(x) = -2x + 4$.Substituting these values into the formula: $x^{3} - 3x^{2} + x + 2 = s(x) \cdot (x - 2) + (-2x + 4)$.Rearranging to solve for $s(x) \cdot (x - 2)$: $s(x) \cdot (x - 2) = (x^{3} - 3x^{2} + x + 2) - (-2x + 4)$.$s(x) \cdot (x - 2) = x^{3} - 3x^{2} + x + 2 + 2x - 4$.$s(x) \cdot (x - 2) = x^{3} - 3x^{2} + 3x - 2$.Now, divide $x^{3} - 3x^{2} + 3x - 2$ by $(x - 2)$ to find $s(x)$.Using polynomial division: $(x^{3} - 2x^{2}) - (x^{2} - 2x) + (x - 2) = x^{2}(x - 2) - x(x - 2) + 1(x - 2) = (x^{2} - x + 1)(x - 2)$.Therefore, $s(x) = x^{2} - x + 1$.

If the polynomial $p(x) = x^{3} - 3x^{2} + x + 2$ is divided by the divisor polynomial $s(x)$, then the quotient polynomial $q(x) = x - 2$ and the remainder polynomial $r(x) = -2x + 4$ are obtained. Find the divisor polynomial $s(x)$.

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