Question

Please help me with my question

Answer 1

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Given ,

`p(x) = ax {}^(3) + 8x {}^(2) + bx + 5 ...(1)`

Also According to Question ,

on dividing polynomial p(x) by ( x + 2 ) , we obtain ( -25 ) as a remainder.

Therefore , putting the value of x as ( x = -2 )

`p( - 2) = a( - 2) {}^(3) + 8( - 2)^(2) + b( -2) + 5`

`p( - 2) = - 8a + 32 - 2b + 5 = - 25 \\ = > - 8a - 2b + 37 = - 25 \\ = > - (8a + 2b) = - (25 + 37) \\ cancelling \: negative \: sign \: at \: both \: sides \\ 8a + 2b = 52 \\ dividing \:both \: sides \: by \: 2 \\ 4a + b = 26...(2)`

also given that..

when the polynomial p(x) is divided by 2x-1 , it gives 0 as the remainder , i.e. , 2x-1 is a root of p(x)

putting value of x = 1/2 in equation (1)

`p( (1)/(2) ) = a( (1)/(2)) {}^(3) + 8( (1)/(2) ) {}^(2) + b( (1)/(2) ) + 5 = 0 \\ = > (a)/(8) + (8)/(4) + (b)/(2) + 5 = 0 \\ = >taking \: LCM \\ = > (a + 16 + 4b + 40)/(8) = 0 \\ = > (a + 4b + 56)/(8) = 0 \\ = > a + 4b = - 56...(3)`

Now , we have equations (2) and

(3) as...

4a + b = 26... (2)

a + 4b = -56... (3)

multiply equation (2) by 4..

16a + 4b = 104...(4)

Solving equations (3) and (4)

a ‎ ‎ ‎ ‎ ‎+ 4b = -56

16a + 4b = 104

-‎ ‎ ‎ ‎ ‎‎ -‎ ‎ ‎ ‎ ‎- ‎ ‎ ‎ ‎ ‎‎ ( bc signs are same )

solving these equations we get,

-15a = -160

a = -160/-15

a = 32/3

putting value of a in (2) , we get...

`4a + b = 23 \\ 4( (32)/(3) ) + b = 23 \\ (128)/(3) + b = 23 \\ b = 23 - (128)/(3) \\ taking \: LCM \\ b = (69 - 128)/(3) \\ = > b = ( -59)/(3)`

hope helpful~