find the value of a and b so thatbthe polynomial f(x)=2x3+ax2 +bx+6 is divisible by x2+x-2
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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....
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