How do you divide #(2x^3 - 11x^2 + 12x + 9)/(x-3)#?

2 Answers
Jul 27, 2018

The remainder is #0# and the quotient is #=2x^2-5x-3#

Explanation:

Let's perform the synthetic division

#color(white)(aaaa)##3##|##color(white)(aaaa)##2##color(white)(aaaa)##-11##color(white)(aaaaaa)##12##color(white)(aaaaa)##9#

#color(white)(aaaaa)##|##color(white)(aaaa)##color(white)(aaaaaaa)##6##color(white)(aaaaa)##-15##color(white)(aaa)##-9#

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

#color(white)(aaaaa)##|##color(white)(aaaa)##2##color(white)(aaaaa)##-5##color(white)(aaaaa)##-3##color(white)(aaaaa)##color(red)(0)#

The remainder is #0# and the quotient is #=2x^2-5x-3#

#(2x^3-11x^2+12x+9)/(x-3)=2x^2-5x-3#

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)=2x^3-11x^2+12x+9#

Therefore,

#f(3)=2*3^3-11*3^2+12*3+9#

#=54-99+36+9#

#=0#

The remainder is #=0#

Jul 27, 2018

#(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)#

Explanation:

#(2x^3-11x^2+12x+9)div(x-3)#

We can divide this polynomial by using synthetic division

We have , #p(x)=2x^3-11x^2+12x+9 and "divisor :"x=3#

We take ,coefficients of #p(x) to 2,-11,12 ,9#

. #3 |# #2color(white)(....)-11color(white)(.......)12color(white)(.......)9#
#ulcolor(white)()|# #ul(0color(white)( ..........)6color(white)(....)-15color(white)(...)-9#
#color(white)(......)2color(white)(......)-5color(white)(...)color(white)(..)-3color(white)(.....)color(violet)(ul|0|#
We can see that , quotient polynomial :

#q(x)=2x^2-5x-3 and"the Remainder"=0#

Hence ,

#(2x^3-11x^2+12x+9)=(x-3)(2x^2-5x-3)+(0)#