Question

the value of a and b so that -x3+ax2-x+b is divisible by (x+1) and (x-2) are:​

Answer 1

here is your answet mate...

Step-by-step explanation:

Given,

f(x)=2x³+ax²+bx+6

we know, x²+x-2 is a factor

i. e. x²-x+2x-2

=x(x-1) +2(x-1)

=(x+2)(x-1) is a factor.

By remainder theorem,

f(1)=0

2(1)³+a(1)²+b*1+6=0

2+a+b+6=0

a+b=-8 ...(i)

Also,

f(-2)=0

2(-2)³+a(-2)²+b(-2)+6=0

-16+4a-2b+6=0

4a-2b=10 ....(ii)

Mutiplying (i) by 2, we get

2a+2b=-16

Adding (ii) , + 4a-2b=10

~~~~~~~~~~~~

6a = -6

a=-6/6=-1

Putting a=-1 in (i) ,

a+b=-8

-1+b=-8

b=-8+1

b=-7

Hope this helps you....

plz plz plz mark me as brainliest plzz...