How do you divide #(-4x^3-15x^2-4x-12)/(x-4) #?

1 Answer
Jul 20, 2018

The remainder is #=-524# and the quotient is #=-4x^2-31x-128#

Explanation:

Let's perform the synthetic division

#color(white)(aaaa)##4##|##color(white)(aaaa)##-4##color(white)(aaaa)##-15##color(white)(aaaaaa)##-4##color(white)(aaaaa)##-12#

#color(white)(aaaaa)##|##color(white)(aaaa)##color(white)(aaaaaaa)##-16##color(white)(aaaa)##-124##color(white)(aaaa)##-512#

#color(white)(aaaaaaaaa)###_________

#color(white)(aaaaaa)##|##color(white)(aaaa)##-4##color(white)(aaaa)##-31##color(white)(aaaa)##-128##color(white)(aaaa)##color(red)(-524)#

The remainder is #=-524# and the quotient is #=-4x^2-31x-128#

ALSO,

Apply the remainder theorem

When a polynomial #f(x)# is divided by #(x-c)#, we get

#f(x)=(x-c)q(x)+r#

Let #x=c#

Then,

#f(c)=0+r#

Here,

#f(x)=-4x^3-15x^2-4x-12#

Therefore,

#f(4)=-4*4^3-15*4^2-4*4-12#

#=-256-240-16-12#

#=-524#

The remainder is #=-524#