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

3 Answers
Phillip E.
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

Jim H
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.)

Sidharth
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

Related questions
See all questions in Long Division of Polynomials
Impact of this question
2140 views around the world
You can reuse this answer
Creative Commons License