I believe you use the remainder theorem for this, which states that the remainder when a polynomial f(x) is divided by (x-k) is
f(k)
Thus the remainder would be f(-1) = 11
2006-11-09 18:11:57
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answer #1
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answered by z_o_r_r_o 6
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The remainder theorem:
When P(x) is divided by x-a, the remainder is P(a)
The polynomial P(x) is being divided by x+1, or x-(-1). The remainder is P(-1):
(-1)^3 + 3(-1)^2 - 2(-1) + 7
-1 + 3 + 2 + 7
11
The remainder is 11.
2006-11-10 02:16:17
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answer #2
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answered by Anonymous
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The remainder = 11
2006-11-10 05:22:46
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answer #3
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answered by Gunner 2
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x3 + 3x2 -2x + 7 : x +1 = x2 + 2x - 4 with remainder 11
2006-11-10 02:15:08
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answer #4
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answered by Rina 2
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(X^3 + 3X^2 - 2X + 7) / (X+1)= X^2 + 2X - 4 + (11)/(X + 1)
which makes 11 the remainder
2006-11-10 04:19:02
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answer #5
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answered by sweetdreams 2
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x^2 + 2x -4
x+1l x^3 + 3x^2 - 2x + 7
-x^3 - x^2
----------------------------
0 + 2x^2 - 2x
-2x^2 -2x
----------------------------
0 -4x + 7
-4x - 4
----------------------------------
0 + 11
Thus your remainder is 11, but whenever you have a remainder, and when you are deviding by 2 tems with a variable, you make sure you take the remainder and put it over the term.
So in this case, the answer or the remainder would be
11 / (x+1)
2006-11-10 04:29:30
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answer #6
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answered by aplpie 3
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Remainder theorem says you sub the value of x that makes the divisor 0, so sub x = -1 and you have the answer.
2006-11-10 02:14:43
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answer #7
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answered by Hy 7
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simple enough, use the Remainder Theorem. or if you're really into math, you can use Synthetic Division to directly find the remainder.
2006-11-10 04:58:09
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answer #8
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answered by JoseABDris 2
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x^2 + 2x - 4 + [11/(x+1)]
_________________
x+1 | x^3 + 3x^2 - 2x +7
-[x^3 + x^2]
2x^2 - 2x
-[2x^2 +2x]
-4x + 7
-[-4x - 4]
11
So your answer is
x^2+2x-4+[11/(x+1)] with the last part being your remainder
2006-11-10 02:31:27
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answer #9
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answered by bourqueno77 4
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x2 2x
x+1 x3 + 3x2 -2x +7
x3 x2
2x2 -2x
2x2 +2x
7
the answer is x2+2x + [7/(x+1)]
2006-11-10 02:17:06
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answer #10
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answered by igot4onit 2
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