Tag: common factors and hcf

Factors of the given numbers are,

Let's look at products of $6$
$6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90$
The pairs whose sums equal to $90$ are:
$6 + 84=90$
$12 + 78=90$
$18 + 72=90$
$66 + 24=90$
$60 + 30=90$
$54 + 36=90$
$48 + 42=90$
Total number of pairs are $7$.
But there can be  $7$ more pairs when the numbers are reversed.
$\therefore$ total number of ordered pair are $14$

$x= 2^3 \times 3 \times 5^2$
$y = 2^2 \times 3^3$
HCF (x,) $=2^2 \times 3$= 12

Pair in Option C is not possible.
because in option C pair is $(42,80)$
Since, Product of numbers $=42\times 80 =3360$
Option C is correct.

Since, $(a + b -c)^6 = (a + b -c)^4 \times (a + b -c)^2 $
$\therefore$  G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ = $(a + b -c)^4$
$\because (a + b -c)^4$ is greatest common in $(a + b -c)^4$ and $(a + b -c)^6$.
Option D is correct.

Let, $p(x) = 20x^2-9x+1$ and $q(x) = 5x^2-6x+1$
$p(x) = 20x^2-9x+1$
         $=20x^2-5x-4x+1$
         $=5x(4x-1)-1(4x-1)$
         $=(5x-1)(4x-1)$
and
$q(x) = 5x^2-6x+1$
         $=5x^2-5x-x+1$
         $=5x(x-1)-1(x-1)$
         $=(5x-1)(x-1)$
$\therefore$ G.C.D of $p(x)$  and  $q(x)=(5x-1)$.
Option B is correct.

Let $p(x) = x^3+x^2-x-1$ and $ q(x) = x^2-1$
$p(x) = x^3+x^2-x-1$
         $ = x^2(x+1)-1(x+1) $
         $ = (x^2-1) (x+1) $
         $ = (x-1)(x+1)(x+1) $
and
$ q(x) = x^2-1$
         $= (x+1)(x-1) $
$\therefore $ G.C.D of $p(x)$ and $q(x)$ =$ (x+1)(x-1) = x^2 - 1 $
Option A is correct.

$p(x) = (x^2-9)(x-3)$
        $= (x^2-3^2)(x-3)$
        $= (x-3)(x+3)(x-3)$
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