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

Explanation:

Step1: Perform polynomial long - division

We divide the polynomial $2x^{4}-3x^{3}-3x^{2}+7x - 3$ by $x^{2}-2x + 1$. First, divide the leading term of the dividend $2x^{4}-3x^{3}-3x^{2}+7x - 3$ (which is $2x^{4}$) by the leading term of the divisor $x^{2}-2x + 1$ (which is $x^{2}$). The result is $2x^{2}$. Multiply $x^{2}-2x + 1$ by $2x^{2}$: $2x^{2}(x^{2}-2x + 1)=2x^{4}-4x^{3}+2x^{2}$. Subtract this from the dividend: $(2x^{4}-3x^{3}-3x^{2}+7x - 3)-(2x^{4}-4x^{3}+2x^{2})=2x^{4}-3x^{3}-3x^{2}+7x - 3 - 2x^{4}+4x^{3}-2x^{2}=x^{3}-5x^{2}+7x - 3$.

Step2: Continue the long - division

Divide the leading term of the new dividend $x^{3}-5x^{2}+7x - 3$ (which is $x^{3}$) by the leading term of the divisor $x^{2}$ to get $x$. Multiply $x^{2}-2x + 1$ by $x$: $x(x^{2}-2x + 1)=x^{3}-2x^{2}+x$. Subtract this from the new dividend: $(x^{3}-5x^{2}+7x - 3)-(x^{3}-2x^{2}+x)=x^{3}-5x^{2}+7x - 3 - x^{3}+2x^{2}-x=-3x^{2}+6x - 3$.

Step3: Final long - division step

Divide the leading term of the new dividend $-3x^{2}+6x - 3$ (which is $-3x^{2}$) by the leading term of the divisor $x^{2}$ to get $-3$. Multiply $x^{2}-2x + 1$ by $-3$: $-3(x^{2}-2x + 1)=-3x^{2}+6x - 3$. Subtract this from the new dividend: $(-3x^{2}+6x - 3)-(-3x^{2}+6x - 3)=0$.

The quotient of $\frac{2x^{4}-3x^{3}-3x^{2}+7x - 3}{x^{2}-2x + 1}$ is $2x^{2}+x - 3$.

Answer:

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