Answer:
[tex]\huge\color{Black}\boxed{\colorbox{pink}{Answer♡}}[/tex]
Given ,
[tex]p(x) = ax {}^{3} + 8x {}^{2} + bx + 5 ...(1)[/tex]
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 )
[tex]p( - 2) = a( - 2) {}^{3} + 8( - 2)^{2} + b( -2) + 5[/tex]
[tex]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)[/tex]
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)
[tex]p( \frac{1}{2} ) = a( \frac{1}{2}) {}^{3} + 8( \frac{1}{2} ) {}^{2} + b( \frac{1}{2} ) + 5 = 0 \\ = > \frac{a}{8} + \frac{8}{4} + \frac{b}{2} + 5 = 0 \\ = >taking \: LCM \\ = > \frac{a + 16 + 4b + 40}{8} = 0 \\ = > \frac{a + 4b + 56}{8} = 0 \\ = > a + 4b = - 56...(3)[/tex]
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...
[tex]4a + b = 23 \\ 4( \frac{32}{3} ) + b = 23 \\ \frac{128}{3} + b = 23 \\ b = 23 - \frac{128}{3} \\ taking \: LCM \\ b = \frac{69 - 128}{3} \\ = > b = \frac{ -59}{3}[/tex]
hope helpful~
