Question

what must be subtracted from x^3-6x^2-15x+80 so that the result is exactly divisible by x^2 +x-12

Answer 1

Hope it helps you✌✌✌✌.....

Answer 2

Answer:

Let ax + b be subtracted from p(x)=x  

3

−6x  

2

−15x+80

so that it is exactly divisible by x  

2

+x−12.

∴s(x)=x  

3

−6x  

2

−15x+80−(ax+b)

=x  

3

−6x  

2

−(15+a)x+(80−b)

Dividend = Divisor × quotient + remainder

But remainder will be zero.

∴ Dividend = Divisor × quotient

⇒s(x)=(x  

2

+x−12)×quotient

⇒s(x)=x  

3

−6x  

2

−(15+a)x+(80−b)

Hence, x(4−a)+(−4−b)=0,x



=0

⇒4−a=0 & (−4−b)=0

⇒a=4 and b=−4

Hence, if in p(x) we subtract 4x - 4 then it is exactly divisible by x  

2

+x−12