Tag: highest common factor (h.c.f.)

$p(x) = (x^2-9)(x-3)$
        $= (x^2-3^2)(x-3)$
        $= (x-3)(x+3)(x-3)$
and
$q(x) = x^2+6x+9$
        $ = x^2+3x+3x+9$
        $ = x(x+3)+3(x+3)$
        $= (x+3)(x+3) $
$\therefore $ G.C.D of $p(x)$ and $q(x) = x+3 $
Option C is correct.

$p(x) =8(x^4-16)$
   $= 4\times 2 [(x^2)^2 - 4^2] $
   $ = 4\times 2 (x^2 - 4) (x^2 + 4) $
   $ = 4\times 2  (x+2)(x-2) (x^2 +4) $
and
 $q(x) = 12(x^3-8)$
       $=4\times 3 (x^3 - 2^3) $
         $=4\times 3 (x - 2)(x^2 + 2x + 4) $
$\therefore $ G.C.D of $p(x)$ and $q(x) = 4(x-2) $
Option A is correct.

${ 3 }^{ 5 }$ = 3x3x3x3x3

Let the two numbers be $a$ and $b$.
$a=quotient\times dividend$
$a=21\times 9$
  $=189$
$\therefore b=\frac { GCD\times LCM }{ a }$
                  $=\frac { 27\times 2079 }{ 189 }$
                  $=297$

$37-1=36, 50-2=48$
$123-3=120$
HCF of 36, 48 and $120=12$
$\therefore$ required number $=12$.

$1134 = 2\times3^{4}\times7$

$10=5\times 2$
$100=5\times 2\times 5\times 2=5^2\times 2^2$
$HCF = 5\times 2=10$

The factors of the given numbers are :
$45  = 1,3,5,9,15$ and $45$
$135 = 1,3,5,9,15,45$ and $135$
$270 = 1,3,5,9,15,45,90,135$ and $270$
The common factors in each of the above numbers are $3,3$ and $5$.
Hence, the GCF is $3\times 3\times 5$ = 45
and as $45$ is also a factor of both $135$ and $270$.
The GCF of $45, 135$ and $270$ is $45$.
The correct answer is Option E, the number $45$.

Ans: E

We know, $42=2\times 3\times 7$

$4$ positive integers are factors of $28$ and $42$ both.
For example,
$1,4, 7, 6$ and $28$ itself is a factor of $28$.
$1,4, 7, 6$ and $42$ itself is factor of $42$.