Question

solve both............

Answer 1

30](5m+1)(5m+3)(5m+4)/5 Write the 12 as 10+2 So the fraction part when dividing by 5 was so the  remainder had to be 2

29] If the HCF of 408 and 1032 can be written as 

By Euclid 's division algorithm,

1032 = 408×2 + 216

408 = 216×1 + 192

216 = 192×1 + 24

192 = 24×8 + 0

Since the remainder becomes 0 here, so HCF of 408 and 1032 is 24

Now, 

1032p - 408*5 = HCF of these numbers

=> 1032p - 2040 = 24

=> 1032p = 24+2040

=> p = 2064/1032

=> p = 2

So value of p is 2.

 

Answer 2

Hiiiii........friend

The answer is here,

$= > \: 408 = {2}^{3} \times 3 \times 17$
$= > \: 1032 = {2}^{3} \times 3 \times 43$

HCF of the numbers,

$= \: {2}^{3} \times 3$

=> 24.

The HCF is in the form => 1032p-408×5 .

=> 1032p-2040 =24.

=> 1032p=2040+24 = 2064.

=> p =2064/1032.

=> p =2.


30) The remainder of (5m+1)(5m+3)(5m+4)/5 is a natural number.

There are two Methods to find remainder .

1st Method:-

Remainder of (5m+1)(5m+3)(5m+4)/5 is,


=> Rem of (5m+1)/5 ×Rem of (5m+3)/5×Rem of (5m+4)/5.

=> 1×3×4.

=> 12.

12 is greater than the divisor (5).
So, Again Divide it by 5.

=> 12/5.

Remainder => 2.

2nd Method:-

=> (5m+1)(5m+3)(5m+4)

$= > \: (25 {m}^{2} + 20m + 3)(5m + 4)$

$= > \: 125 {m}^{3} + 200 {m}^{2} + 95m + 12$

$= > \: 125 {m}^{3} + 200 {m}^{2} + 95m + 10 + 2$

$= > \: 5(25 {m}^{3} + 40 {m}^{2} + 19m + 2) + 2$

So, If we divide the polynomial we, Get 2 as remainder.

Verification:-

Substitute m =1.
=> (5×1+1)(5×1+3)(5×1+4)

=> 6×8×9.

=> 432.

If we divide 432 , We get remainder 2.

Hence The remainder is 2.


:-)Hope it helps u.