Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases:
\( p(x)=x^{3}-6 x^{2}+11 x-6, x=1,2,3 \)
Given:
\( p(x)=x^{3}-6 x^{2}+11 x-6, x=1,2,3 \)
To do:
We have to find whether the indicated numbers are zeros of the polynomials corresponding to them.
Solution:
To find whether $x=1,2,3$ are zeroes of $p(x)$ we have to check if $p(1)=0, p(2)=0$ and $p(3)=0$.
Therefore,
$p(1)=(1)^{3}-6(1)^{2}+11(1)-6$
$=1-6 \times 1+11 \times 1-6$
$=1-6+11-6$
$=12-12$
$=0$
$p(2)=(2)^{3}-6(2)^{2}+11 \times 2-6$
$=8-6 \times 4+22-6$
$=8-24+22-6$
$=30-30$
$=0$
$p(3)=(3)^{3}-6(3)^{2}+11 \times 3-6$
$=27-6 \times 9+33-6$
$=27-54+33-6$
$=60-60$
$=0$
Therefore, $x=1, 2, 3$ are zeroes of $p(x)$.
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