Using suitable identity, evaluate the following:
$999^{2}$
$990001$
$998001$
$999001$
$999999$
$(999)^{2}=(1000-1)^{2}=(1000)^{2}-2 \times(1000) \times 1+1^{2}$
$=100000-2000+1$
$=998001$
Factorise
$25 x^{2}+25 x+6$
By Remainder Theorem find the remainder, when $p(x)$ is divided by $g(x),$ where
$p(x)=x^{3}-2 x^{2}-4 x-1, \quad g(x)=x+1$
For $p(x)=x^{3}+9 x^{2}+26 x+24$ $p(-2)=\ldots \ldots \ldots$
On dividing $p(x)=3 x^{3}-6 x^{2}+5 x-10$ by $(x-2),$ find the remainder.
Factorise the following:
$9 x^{2}-12 x+4$
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