Question

solve it plz answer it

Answer 1

x+2 is factor of ax³+bx²+x-2=0.......(1)
then, x+2 = 0
x = -2 substitute in (1)

a(-2)³+b(-2)²+(-2)-2 = 0
-8a+4b-2-2=0
-8a+4b-4 = 0
-8a+4b = 4
8a = 4b-4
a=4(b-1)/8
a=(b-1)/2..........(2)

when ax³+bx²+x-2= 0 divided by x-2 gives remainder as 4
so, x-2= 0
x=2 substitute in (1)

a(2)³+b(2)²+2-2=4
8a+4b=4
4(2a+b)=4
2a+b=1.........(3)

substitute (2) in (3)

2(b-1)/2+b=1
b-1+b=1
2b=2
b=1 substitute in (2)

a=(1-1)/2
a=0

then the value of a is 0 and b is 1
then the polynomial is (0)x³+1x²+x-2 = 0
becomes x²+x-2=0