Verified Answer :

frika

Here you have two polynomials:

1. The dividend - \(f(x)=2x^4 - 3x^3 - 3x^2 + 7x - 3\)

2. The divisor - \(g(x)=x^2 - 2x + 1\).

Since divisor is perfect square \(g(x)=x^2 - 2x + 1=(x-1)^2\), you should check what is the quotient after division f(x) by (x-1):

\(f(x)=2x^4 - 3x^3 - 3x^2 + 7x - 3=(x-1)(2x^3-x^2-4x+3)=(x-1)(x-1)(2x^2+x-3)=(x-1)^2(2x^2+x-3)=g(x)(2x^2+x-3).\)

Then the quotient is \(2x^2+x-3\).

isyllus

Answer:

Quotient: \(2x^2+x-3\)

Please see the attachment.

Step-by-step explanation:

Given: \((2x^4-3x^3-3x^2+7x-3)\div (x^2-2x+1)\)

We are given rational expression and need to find quotient.

Using long division method to find the quotient.

First we get rid of \(2x^4\) by \(x^2\)

\(x^2-2x+1\) ) \(2x^4-3x^3-3x^2+7x-3\) ( \(2x^2+x-3\)

                 \(-2x^4+4x^3-2x^2\)

                              \(x^3-5x^2+7x\)

                               \(-x^3+2x^2-x\)

                                      \(-3x^2+6x-3\)

                                      \(3x^2-6x+3\)

                                              \(0\)

Hence, The quotient of division is \(2x^2+x-3\)

View image isyllus

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