Aspire Faculty ID #10669 · Topic: NIMCET 2021 · Just now
NIMCET 2021

For what value of p, the polynomial  $x^4-3x^3+2px^2-6$ is exactly divisible by $(x-1)$

Solution

Let the given polynomial be: 
$P(x) = x^4 - 3x^3 + 2px^2 - 6$ 
By the Factor Theorem, if $P(x)$ is exactly divisible by $(x - 1)$, then: 

$P(1) = 0$ 

Substitute $x = 1$ into $P(x)$: 
$P(1) = (1)^4 - 3(1)^3 + 2p(1)^2 - 6 = 0$ 
$1 - 3 + 2p - 6 = 0$ 
$2p - 8 = 0$ 
$2p = 8$ 
$p = 4$ 
 Thus, the value of $p$ is $4$.
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