Question

use the long division method to find the result when $2x^4 + 3x^3 + 2x^2 + 15x + 14$ is divided by $2x^2 - 3x + 7$. if there is a remainder, express the result in the form $q(x) + \frac{r(x)}{b(x)}$.

Explanation:

Step1: Divide leading terms

$\frac{2x^4}{2x^2} = x^2$

Step2: Multiply divisor by $x^2$

$x^2(2x^2 - 3x + 7) = 2x^4 - 3x^3 + 7x^2$

Step3: Subtract from dividend

$(2x^4 + 3x^3 + 2x^2 + 15x + 14) - (2x^4 - 3x^3 + 7x^2) = 6x^3 - 5x^2 + 15x + 14$

Step4: Divide new leading terms

$\frac{6x^3}{2x^2} = 3x$

Step5: Multiply divisor by $3x$

$3x(2x^2 - 3x + 7) = 6x^3 - 9x^2 + 21x$

Step6: Subtract from new polynomial

$(6x^3 - 5x^2 + 15x + 14) - (6x^3 - 9x^2 + 21x) = 4x^2 - 6x + 14$

Step7: Divide new leading terms

$\frac{4x^2}{2x^2} = -2$

Step8: Multiply divisor by $-2$

$-2(2x^2 - 3x + 7) = -4x^2 + 6x - 14$

Step9: Subtract to find remainder

$(4x^2 - 6x + 14) - (-4x^2 + 6x - 14) = 8x + 28$

Step10: Combine quotient and remainder

Quotient $q(x)=x^2+3x-2$, remainder $r(x)=8x+28$, divisor $b(x)=2x^2-3x+7$

Answer:

$x^2 + 3x - 2 + \frac{8x + 28}{2x^2 - 3x + 7}$