How do you divide #-d^3 + 7d^2 - 11d - 3# by #d+3#?

1 Answer
Aug 5, 2015

This can't be done exactly. I didn't work it out below, but if remainders are accepted, then the answer is -d^2 + 10d - 41 with a remainder of 120.

Explanation:

To do this, the expression needs to be able to be factored into something and (d+3).

Honestly, with polynomials of great length like this one, it's best to play around with the equation and see what you can get. Let's try to make that #-d^3# term first.

We can multiply (d+3) by #-d^2# to get #-d^3 - 3d^2#

Now, let's get the next term to be what we want it to be. It should be #7d^2#, and right now it's# -3d^2#, so we need to add #10 d^2#.

#-d^2(d+3) + 10d(d+3) = -d^3 - 3d^2 + 10d^2 + 30d = -d^3 + 7d^2 + 30d#

Last step - make the 3rd term right and see if the last term automatically matches up. If it doesn't then this polynomial can't be divided evenly.

#(-d^2+10d)(d+3) - 41(d+3) = -d^3 + 7d^2 - 11d - 123#

Whoa... that last term isn't right. That means this polynomial isn't divisible by (d+3). You can check this with an online polynomial divider or factorer. You'll find that this gives a remainder and the polynomial doesn't have (d+3) as a factor.