If the polynomial#x^4-6x^3+16x^2-25x+10# is divided by #x^2-2x+k# the remainder is #x+a# find #k# and #a#.?

2 Answers
Cem Sentin
Nov 25, 2017

#k=5# and #a=-5#

Explanation:

From #x^2-2x+k=0#, #x^2=2x-k#

After replacing #x^2# into #2x-k# in polynomial and equating remainder into #x+a#,

#(x^2)^2-6x*(2x-k)+16*(2x-k)-25x+10=x+a#

#(2x-k)^2-12x^2+6kx+32x-16k-25x+10=x+a#

#(2x-k)^2-12x^2+6kx+7x-16k+10=x+a#

#4x^2-4xk+k^2-12x^2+(6k+7)*x-16k+10=x+a#

#-8x^2-4xk+k^2+(6k+7)*x-16k+10=x+a#

#-8*(2x-k)-4xk+k^2+(6k+7)*x-16k+10=x+a#

#-16x+8k+(2k+7)*x+k^2-16k+10=x+a#

#(2k-9)*x+k^2-8k+10=x+a#

After equating coefficients,

#2k-9=1# and #a=k^2-8k+10#

Hence #k=5# and #a=-5#

Ratnaker Mehta
Nov 27, 2017

# k=5, and, a=-5.#

Explanation:

The polynomial #p(x)=x^4-6x^3+16x^2-25x+10,# when

divided by #(x^2-2x+k),# leaves the remainder # r(x)=(x+a).#

If the quotient polynomial is #q(x),# then, knowing that,

#p(x)=(x^2-2x+k)q(x)+r(x),# we have,

#p(x)-r(x)=(x^2-2x+k)q(x).#

This means that, #p(x)-r(x)# is divisible by #(x^2-2x+k).#

So, when #p(x)-q(x)# is divided by #(x^2-2x+k),# the Remainder

must be #0.#

Now,

#p(x)-r(x)=(x^4-6x^3+16x^2-25x+10)-(x+a), i.e.,#

#p(x)-r(x)=x^4-6x^3+16x^2-26x+(10-a).#

By Long Division of #p(x)-r(x)# by #(x^2-2x+k),# we get,

#p(x)-r(x)=x^4-6x^3+16x^2-26x+(10-a),#

#=(x^2-2x+k){x^2-4x+(8-k)}+{2(k-5)x+k^2-8k+10-a}.#

#:."The Remainder="0 rArr {2(k-5)x+k^2-8k+10-a}=0.#

#:. (k-5)=0, and, k^2-8k+10-a=0.#

#:. k=5, and, a=k^2-8k+10=25-40+10=-5.#

Enjoy Maths.!

Related questions
Impact of this question
2180 views around the world
You can reuse this answer
Creative Commons License