Question

the polynomial 6 x^4+ 8 x cube + 17 X square + 21 X + 7 is divided by another polynomial 3 X square + 4 x + 1 then remainder comes out to be a x + b find a and b

Answer 1

Hello Mate!

When ,

( 6x^4 + 8x^3 + 17x^2 + 21x + 7 ) ÷ ( 3x^2 + 4x + 1 )
Then, remainder is ( ax + b ), as given in question.

Now, on factoring 3x^2 + 4x + 1 we get
3x^2 + 3x + x + 1
3x( x + 1 ) + 1 ( x + 1 )
( 3x + 1 )( x + 1 ) = 0
3x + 1 = 0 and x + 1 = 0
3x = - 1 and x = - 1
x = - 1/3 or - 1


So value of x are - 1/3 and - 1 respectively

Now, if we subtract remainder in dividend then it will be completely divided by divisor.

p( x ) = 6x^4 + 8x^3 + 17x^2 + 21x + 7 - ax - b
0 = 6(-1)^4 + 8(-1)^3 + 17(-1)^2 + 21(-1) + 7 - a(-1) - b
0 = 6 - 8 + 17 - 21 + 7 + a - b
0 = 1 + a - b ---------(1)

0 = 6(-1/3)^4 + 8(-1/3)^3 + 17(-1/3)^2 + 21(-1/3) + 7 - a(-1/3) - b
0 = 2/27 - 8/27 +17/9 - 7 + 7 + 1a/3 - b
0 = ( 2 - 8 + 51 + 9a - 27b )/ 27
0 = 2 - 8 + 51 - 9a + 27b
0 = 45 + 9a - 27b -------(2)

Multiply - 9 from equation (1), then it becomes

0 = -9 - 9a + 9b ------(3)

Now, add (2) and (3)

0 = 36 - 18b
- 36 = -18b
2 = b

0 = 1 + a - b
- 1 = a - 2
- 1 + 2 = a
1 = a

Verification, refer to image

Hope it helps☺!✌

Answer 2

Answer:

I am not understand your question and not give answer