For what value of p, the polynomial \(x^4-3x^3+2px^2-6\) is exactly divisible by \((x-1)\)
We know that when \(f(x)\) is divided by \((x - a)\), the remainder is \(f(a)\). To be exactly divisible by \((x - 1)\), the value of the polynomial at \(x = 1\) must be 0. Hence: \[ 1^4 - 3(1^3) + 2p(1^2) - 6 = 0 \implies p = 4 \]
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