How to solve?

Question

#(340x^4-x^3-x^2-x-1)/(4x-1)#

466 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-10T23:40:40

See below.

We need #Q(x), R(x)# such that

#340x^4-x^3-x^2-x-1 =(4x-1)Q(x)+R(x)#

Here the remainder #R(x) = r# because the divisor is of degree #1# and #Q(x) = p_3(x) = a x^3+bx^2+c x+d# then we have

#340x^4-x^3-x^2-x-1 =(4x-1)( a x^3+bx^2+c x+d)+r# or

#340x^4-x^3-x^2-x-1 -((4x-1)( a x^3+bx^2+c x+d)+r)=0#

after grouping coefficients we get

#(340-4a)x^4+(a-1-4b)x^3+(b-1-4c)x^2+(c-1-4d)x+d-1-r=0#

and the condition is

#{(340-4a=0),(a-1-4b=0),(b-1-4c=0),(c-1-4d=0),(d-1-r=0):}#

and easily we obtain

#a = 85, b = 21, c=5, d = 1, r = 0#

or

#340x^4-x^3-x^2-x-1 =(4x-1)(85x^3+21x^2+5 x+1)#