If P(x)=2x^3-3x^2-5x+6P(x)=2x3−3x2−5x+6 is divided by x-2x−2, then (1) what is the remainder; (2) what is the quotient and (3) how we can express P(x)P(x) in terms of factors?

2 Answers
Cesareo R.
Mar 27, 2017

See below.

Explanation:

I think some sign or coefficient are not well fitted because

assuming q(x)=ax^2+bx+cq(x)=ax2+bx+c and substituting

2x^3-3x^2-5x+6=(ax^2+bx+c)(x-2)+r2x3−3x2−5x+6=(ax2+bx+c)(x−2)+r or

2x^3-3x^2-5x+6 = ax^3+(b-2a)x^2+(c-2b)x+r-2c2x3−3x2−5x+6=ax3+(b−2a)x2+(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)

Shwetank Mauria
Mar 28, 2017
  1. r=0, P(x)=2x^3-3x^2-5x+6
  2. q(x)=(2x+3)(x-1)
  3. P(x)=(2x+3)(x-1)(x-2)

Explanation:

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)

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