divide $x^{3}-3x^{2}+x - 2enclose{longdiv}{10x^{4}-14x^{3}-10x^{2}+6x - 10}$\nthe quotient is $square…

divide $x^{3}-3x^{2}+x - 2enclose{longdiv}{10x^{4}-14x^{3}-10x^{2}+6x - 10}$\nthe quotient is $square x+square$\nthe remainder is $square x^{2}+square x+square$\ndone
Answer
Explanation:
Step1: Divide leading terms
Divide the leading - term of the dividend $10x^{4}$ by the leading - term of the divisor $x^{3}$. The result is $10x$.
Step2: Multiply and subtract
Multiply the divisor $x^{3}-3x^{2}+x - 2$ by $10x$: $10x(x^{3}-3x^{2}+x - 2)=10x^{4}-30x^{3}+10x^{2}-20x$. Subtract this from the dividend: $(10x^{4}-14x^{3}-10x^{2}+6x - 10)-(10x^{4}-30x^{3}+10x^{2}-20x)=16x^{3}-20x^{2}+26x - 10$.
Step3: Divide leading terms again
Divide the leading - term of the new dividend $16x^{3}$ by the leading - term of the divisor $x^{3}$, the result is $16$.
Step4: Multiply and subtract again
Multiply the divisor $x^{3}-3x^{2}+x - 2$ by $16$: $16(x^{3}-3x^{2}+x - 2)=16x^{3}-48x^{2}+16x - 32$. Subtract this from the new dividend: $(16x^{3}-20x^{2}+26x - 10)-(16x^{3}-48x^{2}+16x - 32)=28x^{2}+10x + 22$.
Answer:
The quotient is $10x + 16$ The remainder is $28x^{2}+10x + 22$