2007-06-23 18:29:08 · 6 answers · asked by matcav2002 1 in Science & Mathematics ➔ Mathematics
x^5(-8+3x^5)
2007-06-23 18:35:03 · answer #1 · answered by Anonymous · 0⤊ 2⤋
This problem can be solved by algebraic division.
When you divide x^4+x^2-80 by x+3, then u get a result of x^3-3x^2+10x-30 and a remainder of 160.
2007-06-24 01:55:47 · answer #2 · answered by gothic_vampy_666 1 · 0⤊ 0⤋
x^4 + x² - 80 = (x + 3)(x³ - 3x² + 10x - 30) + 10
2007-06-24 01:48:56 · answer #3 · answered by fred 5 · 0⤊ 0⤋
if the constant is 90 rather than 80 then
ans x^3 - 3x^2 +10x-30
2007-06-24 01:43:45 · answer #4 · answered by tarun_maths 1 · 0⤊ 0⤋
Use synthetic division:
-3...1...0....1......0.....-80
...........-3...9....-30.....90
......1...-3..10...-30...10
Answer:
quotient = x^3 - 3x^2 + 10x - 30
remainder = 10
2007-06-24 02:22:18 · answer #5 · answered by mathjoe 3 · 0⤊ 0⤋
. . . . x^3 - 3x^2 + 10x - 30x
x + 3)x^4 . . . . . . + x^2 . .. . . - 80
. . . - x^4 - 3x^3
. . . . . . . . - 3x^3 . + x^2 . .. . . - 80
. . . . . . . . . 3x^3 + 9x^2
. . . . . . . . . . . . . . 10x^2 . .. . . - 80
. . . . . . . . . . . . . - 10x^2 - 30x
. . . . . . . . . . . . . . . . . . . - 30x - 80
. . . . . . . . . . . . . . . . . . . . 30x + 90
. . . . . . . . . . . . . . . . . . . . . . . . . 10
x^3 - 3x^2 + 10x - 30x + 10/(x + 3)
2007-06-24 02:00:46 · answer #6 · answered by Helmut 7 · 0⤊ 0⤋