Question

divide 2x6-3x4-5x2+3x-2 by 4-3x-x2

Answer 1

Given:

2x6-3x4-5x2+3x-2, 4-3x-x2

To find:

Divide 2x6 - 3x4 - 5x2 + 3x - 2 by 4 - 3x - x2

Solution:

2x6 - 3x4 - 5x2 + 3x - 2 ÷ 4 - 3x - x2 =

$\dfrac{2x^6-3x^4-5x^2+3x-2}{4-3x-x^2}$

$=-2x^4+\dfrac{-6x^5+5x^4-5x^2+3x-2}{-x^2-3x+4}$

$=-2x^4+6x^3+\dfrac{23x^4-24x^3-5x^2+3x-2}{-x^2-3x+4}$

$=-2x^4+6x^3-23x^2+\dfrac{-93x^3+87x^2+3x-2}{-x^2-3x+4}$

$=-2x^4+6x^3-23x^2+93x+\dfrac{366x^2-369x-2}{-x^2-3x+4}$

$=-2x^4+6x^3-23x^2+93x-366+\dfrac{-1467x+1462}{-x^2-3x+4}$

The quotient is -2x⁴ + 6x³ - 23x² + 93x - 366

The remainder is -1467x + 1462