If (x + 1) is a factor of {x^4} - (p - 3){x^3 | vedclass

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(A) Let $f(x) = {x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$.Since $(x + 1)$ is a factor of $f(x)$,by the Factor Theorem,$f(-1) = 0$.Substitu

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$4$

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(A) Let $f(x) = {x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$.Since $(x + 1)$ is a factor of $f(x)$,by the Factor Theorem,$f(-1) = 0$.Substituting $x = -1$ into the polynomial:$(-1)^4 - (p - 3)(-1)^3 - (3p - 5)(-1)^2 + (2p - 7)(-1) + 6 = 0$$1 - (p - 3)(-1) - (3p - 5)(1) - (2p - 7) + 6 = 0$$1 + (p - 3) - (3p - 5) - 2p + 7 + 6 = 0$$1 + p - 3 - 3p + 5 - 2p + 7 + 6 = 0$Combining like terms:$(1 - 3 + 5 + 7 + 6) + (p - 3p - 2p) = 0$$16 - 4p = 0$$4p = 16$$p = 4$.

If $(x + 1)$ is a factor of ${x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$,then $p = $

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