How do you use polynomial division to determine whether x=3 is a root of #x^3+x^2+15-x#?

1 Answer
Nov 24, 2015

Answer:

Divide #(x^3 + x^2 -x + 15) -: (x-r)# with long polynomial division.

If you can divide without a remainder, #x = r# is a root.
If you have a remainder, it's not a root.

Explanation:

Well, just to check if #x = 3# is a root of your term, it would be easier to plug #x = 3# and see if the result is #0#. :-)

But of course, it is also possible to determine the outcome with polynomial division.

To do so, you must divide #(x^3 + x^2 -x + 15)# by #(x-3)#.

I know that in some countries, a different notation for long division is being used but I will use the one that I'm familiar with and I hope that it will be no problem for you to rewrite it in your notation if necessary.

#color(white)(xx)(x^3 + x^2 color(white)(xx) - x color(white)(xx) + 15) : (x-3) = x^2 + 4x + 11#
#color(white)(i)-(x^3 - 3x^2)#
#color(white)(xxx)color(white)(xxxxxx)/color(white)(x)#
#color(white)(xxxxxxx) 4 x^2 color(white)(x)- x#
#color(white)(xxxxx) -(4 x^2 - 12x)#
#color(white)(xxxxxxx)color(white)(xxxxxxxx)/color(white)(x)#
#color(white)(xxxxxxxxxxxx) 11x color(white)(x) + 15#
#color(white)(xxxxxxxxxx) -(11x color(white)(x) - 33)#
#color(white)(xxxxxxxxxxxx)color(white)(xxxxxxxx)/color(white)(x)#
#color(white)(xxxxxxxxxxxxxxxxxx) 48#

At the end of the computation, there is the remainder #48#.

As you have a remainder, it means that #x = 3# is not a root of #x^3 + x^2 -x + 15#.