use the long division method to find the result when $3x^{4}+2x^{3}-3x^{2}+12x - 6$ is divided by $x^{2}+2x…

use the long division method to find the result when $3x^{4}+2x^{3}-3x^{2}+12x - 6$ is divided by $x^{2}+2x - 1$. if there is a remainder, express the result in the form $q(x)+\frac{r(x)}{b(x)}$.

use the long division method to find the result when $3x^{4}+2x^{3}-3x^{2}+12x - 6$ is divided by $x^{2}+2x - 1$. if there is a remainder, express the result in the form $q(x)+\frac{r(x)}{b(x)}$.

Answer

Explanation:

Step1: Divide the leading terms

Divide the leading - term of the dividend $3x^{4}+2x^{3}-3x^{2}+12x - 6$ (which is $3x^{4}$) by the leading - term of the divisor $x^{2}+2x - 1$ (which is $x^{2}$). We get $3x^{2}$ as the first term of the quotient $q(x)$. $$3x^{4}\div x^{2}=3x^{2}$$

Step2: Multiply and subtract

Multiply the divisor $x^{2}+2x - 1$ by $3x^{2}$: $3x^{2}(x^{2}+2x - 1)=3x^{4}+6x^{3}-3x^{2}$. Subtract this from the dividend: $$(3x^{4}+2x^{3}-3x^{2}+12x - 6)-(3x^{4}+6x^{3}-3x^{2})=-4x^{3}+12x - 6$$

Step3: Repeat the process

Divide the leading - term of the new dividend $-4x^{3}+12x - 6$ (which is $-4x^{3}$) by the leading - term of the divisor $x^{2}$ to get $-4x$ as the next term of the quotient. $$-4x^{3}\div x^{2}=-4x$$ Multiply the divisor $x^{2}+2x - 1$ by $-4x$: $-4x(x^{2}+2x - 1)=-4x^{3}-8x^{2}+4x$. Subtract this from the new dividend: $$(-4x^{3}+12x - 6)-(-4x^{3}-8x^{2}+4x)=8x^{2}+8x - 6$$

Step4: One more time

Divide the leading - term of the new dividend $8x^{2}+8x - 6$ (which is $8x^{2}$) by the leading - term of the divisor $x^{2}$ to get $8$ as the next term of the quotient. $$8x^{2}\div x^{2}=8$$ Multiply the divisor $x^{2}+2x - 1$ by $8$: $8(x^{2}+2x - 1)=8x^{2}+16x - 8$. Subtract this from the new dividend: $$(8x^{2}+8x - 6)-(8x^{2}+16x - 8)=-8x + 2$$

Answer:

$3x^{2}-4x + 8+\frac{-8x + 2}{x^{2}+2x - 1}$