How do you divide # 6x^342+18x^63-5x^2-20# by #x-1#?

3 Answers
Jun 17, 2015

You can't

Explanation:

If #x-1# is a factor then the value of #x=1# will give 0
#6+18-5-20=-1# so #x-1# can't be a factor

Jun 19, 2015

Using synthetic division, it's not so bad. But I don't know how to format division here.

Explanation:

It goes something like this:

#{:(342, 341, 340, " . . . ", 64, 63, 62, " . . . ", 3, 2, 1,"cst"), (6,0,0," . . . ",0,18,0," . . . ",0,-5,0,-20),( color(white)"SS", 6, 6, " . . . ", 6, 6, 24, " . . . ", 24, 24, 19, 19),(6, 6, 6, " . . . ", 6, 24, 24, " . . . ", 24, 19, 19, -1):}#

The quotient is:
#6x^341+6x^340 + 6x^339+* * * +6x^63+24x^62+24x^61 + * * * +24x^2+19x+19#:

The remainder is #-1#.
(Which is nice, because it is easy to see that the expression evaluates at #x=1# to #-1# so that agrees with the remainder, as the Remainder Theorem tells us they will.)

Nov 15, 2015

-1 is the remainder

Explanation:

For this i recommend Euclid lemma

#p(x) = q(x)(x-1) + r(x)#

#p(1) = r(x)#

Now we get that #r(x) = -1#

Now #p(x) = q(x)(x-1) -1#

#=> #we can divide #(p(x) + 1)/( x-1)#

Now #p'(x) = 6x^342+18x^63-5x^2-19#

I believe you question is to find the remainder if not I think you do what Jim is proposing