If #P(x)=2x^3-3x^2-5x+6# is divided by #x-2#, then (1) what is the remainder; (2) what is the quotient and (3) how we can express #P(x)# in terms of factors?

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Answer 1 · 2017-03-27T15:38:21

See below.

I think some sign or coefficient are not well fitted because

assuming #q(x)=ax^2+bx+c# and substituting

# 2x^3-3x^2-5x+6=(ax^2+bx+c)(x-2)+r# or

# 2x^3-3x^2-5x+6 = ax^3+(b-2a)x^2+(c-2b)x+r-2c#

Identifying coefficients,

#{(a=2),(b-2a=-3),(c-2b=5),(r-2c=6):}#

Solving for #a,b,c,r# we get

#a=2,b=1,c=-3,r=0# so

#q(x)=2x^2+x-3# and #r=0#

Solving now #q(x)=0# we obtain

#x =(-3/2,1)# or

#q(x)=2(x+3/2)(x-1)#

Concerning #p(x)= 2x^3-3x^2-5x+6=0#

we know that

#p(x)=q(x)(x-2)#

or

#p(x)=2(x+3/2)(x-1)(x-2)#

Answer 2 · 2017-03-28T11:58:30

#P(x)=2x^3-3x^2-5x+6=q(x)(x-2)+r#

Using remainder theorem #P(2)=r#

and therefore #r=2*2^3-3*2^2-5*2+6#

#=16-12-10+6=0# and

#P(x)=2x^3-3x^2-5x+6=q(x)(x-2)+0#

or #q(x)(x-2)=2x^3-3x^2-5x+6#

#=2x^3-3x^2-5x+6#

#=2x^2(x-2)+x(x-2)-3(x-2)#

#=(2x^2+x-3)(x-2)#

#=(2x^2+3x-2x-3)(x-2)#

#=(x(2x+3)-1(2x+3))(x-2)#

#=(2x+3)(x-1)(x-2)#

and #P(x)=(2x+3)(x-1)(x-2)#