Question

What is the value of b when g(x)=bx³+x²+3 is divided by x+1 the remainder is 7

Answer 1

Answer:

We know that

Dividend=Divisor×Quotient+Remainder

So if we subtract remainder 7 from g(x) then g(x) will be completely divisible by x+1

=bx³+x²+3-7

=bx³+x²-4

Now we get a number which is completely divisible by x+1.This means

x+1=0

x = -1

On putting value of x in bx³+x²-4 we will get value of x

g(x)=bx³+x²-4

g(-1)=0

bx³+x²-4=0

b(-1)³+(-1)²-4=0

-b+1-4=0

1-4=b

-3=b