what is the quotient of (2x^4 - 3x^3 - 3x^2 + 7x - 3)÷(x^2 - 2x + 1)?\n2x^4 - 3x^3-\frac{11}{2}\n2x^2 + x…

Explanation:

Step1: Perform polynomial long - division

We divide the leading term of the dividend $2x^{4}-3x^{3}-3x^{2}+7x - 3$ by the leading term of the divisor $x^{2}-2x + 1$. The leading term of the dividend is $2x^{4}$ and the leading term of the divisor is $x^{2}$. So, the first term of the quotient is $\frac{2x^{4}}{x^{2}}=2x^{2}$.

Step2: Multiply the divisor by the first - term of the quotient

Multiply $x^{2}-2x + 1$ by $2x^{2}$: $2x^{2}(x^{2}-2x + 1)=2x^{4}-4x^{3}+2x^{2}$.

Step3: Subtract the result from the dividend

$(2x^{4}-3x^{3}-3x^{2}+7x - 3)-(2x^{4}-4x^{3}+2x^{2})=2x^{4}-3x^{3}-3x^{2}+7x - 3 - 2x^{4}+4x^{3}-2x^{2}=x^{3}-5x^{2}+7x - 3$.

Step4: Repeat the process

The leading term of the new dividend $x^{3}-5x^{2}+7x - 3$ is $x^{3}$. Dividing $x^{3}$ by $x^{2}$ gives $x$. Multiply $x^{2}-2x + 1$ by $x$: $x(x^{2}-2x + 1)=x^{3}-2x^{2}+x$.

Step5: Subtract again

$(x^{3}-5x^{2}+7x - 3)-(x^{3}-2x^{2}+x)=x^{3}-5x^{2}+7x - 3 - x^{3}+2x^{2}-x=-3x^{2}+6x - 3$.

Step6: One more time

The leading term of $-3x^{2}+6x - 3$ is $-3x^{2}$. Dividing $-3x^{2}$ by $x^{2}$ gives $-3$. Multiply $x^{2}-2x + 1$ by $-3$: $-3(x^{2}-2x + 1)=-3x^{2}+6x - 3$.

Step7: Subtract for the last time

$(-3x^{2}+6x - 3)-(-3x^{2}+6x - 3)=0$.

Answer:

D. $2x^{2}+x - 3$