From the following polynomials,find out which of them has $(x-1)$ as a factor:
$x^{3}+4x^{2}+x-6$
$x^{3}+4x^{2}+x-6$
(A) To determine if $(x-1)$ is a factor of the polynomial $p(x) = x^{3}+4x^{2}+x-6$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$a = 1$.
Substitute $x = 1$ into the polynomial:
$p(1) = (1)^{3} + 4(1)^{2} + (1) - 6$
$p(1) = 1 + 4(1) + 1 - 6$
$p(1) = 1 + 4 + 1 - 6$
$p(1) = 6 - 6 = 0$
Since $p(1) = 0$,$(x-1)$ is a factor of the given polynomial.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$a = 1$.
Substitute $x = 1$ into the polynomial:
$p(1) = (1)^{3} + 4(1)^{2} + (1) - 6$
$p(1) = 1 + 4(1) + 1 - 6$
$p(1) = 1 + 4 + 1 - 6$
$p(1) = 6 - 6 = 0$
Since $p(1) = 0$,$(x-1)$ is a factor of the given polynomial.
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