To find the quotient of (3x^4 – 4x^2 + 8x – 1) ÷ (x – 2),
First: divide 3x^4 by x to get 3x^3, then multiply 3x^3 to x - 2 to get 3x^4 - 6x^3 and subtract 3x^4 - 6x^3 from 3x^4 – 4x^2 + 8x – 1 to get 6x^3 - 4x^2 + 8x - 1
Next: divide 6x^3 by x to get 6x^2, then multiply 6x^2 to x - 2 to get 6x^3 - 12x^2 and subtract 6x^3 - 12x^2 from 6x^3 - 4x^2 + 8x - 1 to get 8x^2 + 8x - 1
Next: divide 8x^2 by x to get 8x, then multiply 8x to x - 2 to get 8x^2 - 16x and subtract 8x^2 - 16x from 8x^2 + 8x - 1 to get 24x - 1
Finally: divide 24x by x to get 24, then multiply 24 to x - 2 to get 24x - 48 and subtract 24x - 48 from 24x - 1 to get 47.
Thus, <span>(3x^4 – 4x^2 + 8x – 1) ÷ (x – 2) = 3x^3 + 6x^2 + 8x + 24 Remainder 47</span>
