When x3+3x2-2x+7 is divided by x+1, the remainder is ?
回答 (8)
2008-12-31 5:21 am
✔ 最佳答案
.............. __x^2+2x-4____ ramainder is 11 answer.......... x+1|x^3+3x^2-2x+7
................ x^3+x^2
................---------------
.................2x^2-2x
.................2x^2+2x
................---------------
....................-4x+7
................... -4x-4
....................------------
.........................11
2008-12-31 4:59 am
= (x³ + 3x² - 2x + 7)/(x + 1)
= x² + 2x - 4 rem 11
Answer: 11 is the remainder.
Proof:
= (x³ + 3x² - 2x + 7) - ([x² + 2x - 4][x + 1])
= (x³ + 3x² - 2x + 7) - (x³ + x² + 2x² + 2x - 4x - 4)
= (x³ + 3x² - 2x + 7) - (x³ + 3x² - 2x - 4)
= x³ + 3x² - 2x + 7 - x³ - 3x² + 2x + 4
= 11
= x² + 2x - 4 rem 11
Answer: 11 is the remainder.
Proof:
= (x³ + 3x² - 2x + 7) - ([x² + 2x - 4][x + 1])
= (x³ + 3x² - 2x + 7) - (x³ + x² + 2x² + 2x - 4x - 4)
= (x³ + 3x² - 2x + 7) - (x³ + 3x² - 2x - 4)
= x³ + 3x² - 2x + 7 - x³ - 3x² + 2x + 4
= 11
2016-10-19 10:18 pm
the rest theorem: while P(x) is divided by skill of x-a, the rest is P(a) The polynomial P(x) is being divided by skill of x+a million, or x-(-a million). the rest is P(-a million): (-a million)^3 + 3(-a million)^2 - 2(-a million) + 7 -a million + 3 + 2 + 7 11 the rest is 11.
2008-12-31 8:13 am
...................x^2 + 2x - 4
......._________________
x + 1)x^3 + 3x^2 - 2x + 7
.........x^3 + x^2
-------------------------------------
.................2x^2 - 2x
.................2x^2 + 2x
-------------------------------------
..........................-4x + 7
..........................-4x - 4
-------------------------------------
.................................11
(x^3 + 3x^2 - 2x + 7)/(x + 1)
= (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
(x^3 + 3x^2 - 2x + 7)/(x + 1)
= (x^2 + 2x - 4) + 11/(x + 1)
2008-12-31 5:05 am
Long Division Method:
….….… x² + 2x - 4
........ --------------------------
x + 1 ) x³ + 3x² - 2x + 7
….… - x³ - 1x²
….… ---------------
….…….… + 2x² - 2x
.….…..… - 2x² - 2x
..….….…. ---------------
…....……..…… - 4x + 7
….…..…...…… + 4x + 4
……........…...... ---------------
….……..…….......… + 11
(x³ + 3x² - 2x + 7) / (x + 1) = x² + 2x - 4 R 11
ANSWER: The remainder is 11.
~~~~~~~~~~~~~~~~~~~~~~
Formula Method:
Given: p(x) / (x - a)
Means: R = p(a)
Given: x³ + 3x² - 2x + 7
Means: p(x) = x³ + 3x² - 2x + 7
Given: divided by (x + 1)
Means: x - -1 ==> a = -1
R = p(a) =
p(-1) = (-1)³ + 3(-1)² - 2(-1) + 7 =
-1 + 3(1) + 2 + 7 =
-1 + 3 + 2 + 7 =
11
ANSWER: The remainder is 11.
….….… x² + 2x - 4
........ --------------------------
x + 1 ) x³ + 3x² - 2x + 7
….… - x³ - 1x²
….… ---------------
….…….… + 2x² - 2x
.….…..… - 2x² - 2x
..….….…. ---------------
…....……..…… - 4x + 7
….…..…...…… + 4x + 4
……........…...... ---------------
….……..…….......… + 11
(x³ + 3x² - 2x + 7) / (x + 1) = x² + 2x - 4 R 11
ANSWER: The remainder is 11.
~~~~~~~~~~~~~~~~~~~~~~
Formula Method:
Given: p(x) / (x - a)
Means: R = p(a)
Given: x³ + 3x² - 2x + 7
Means: p(x) = x³ + 3x² - 2x + 7
Given: divided by (x + 1)
Means: x - -1 ==> a = -1
R = p(a) =
p(-1) = (-1)³ + 3(-1)² - 2(-1) + 7 =
-1 + 3(1) + 2 + 7 =
-1 + 3 + 2 + 7 =
11
ANSWER: The remainder is 11.
2008-12-31 5:01 am
If p(x) is a polynomial, the remainder of p(x)/(x-a) is p(a).
Proof by induction on degree of polynomial.
Consider (rx-a)/(x-b). The result is r with remainder (rx-a) - (rx-rb) = which is rb-a = p(a)
Suppose true for n. Let p be a polynomial of degree n+1. Consider p/(x-a). p = (x-b)p'(x) for some root b of p.
p/(x-a) = (x-b)p'(x)/(x-a) = (x-a)(x-b)q(x) + (x-b)p'(a). Now (x-b)p'(a) is p'(a)(x-a) + a remainder which is p'(a)(x-b -(x-a) = p'(a)(a-b) which is p(a).
So the induction proof is complete.
Proof by induction on degree of polynomial.
Consider (rx-a)/(x-b). The result is r with remainder (rx-a) - (rx-rb) = which is rb-a = p(a)
Suppose true for n. Let p be a polynomial of degree n+1. Consider p/(x-a). p = (x-b)p'(x) for some root b of p.
p/(x-a) = (x-b)p'(x)/(x-a) = (x-a)(x-b)q(x) + (x-b)p'(a). Now (x-b)p'(a) is p'(a)(x-a) + a remainder which is p'(a)(x-b -(x-a) = p'(a)(a-b) which is p(a).
So the induction proof is complete.
2008-12-31 5:01 am
11
2008-12-31 5:07 am
p(x)=x3+3x2-2x+7
by the remainder theorem, p(-1) is the remainder.
p(-1)=( -1)3 + 3(-1)2 - 2(-1) +7
= -1 +3 +2 +7
= 11
by the remainder theorem, p(-1) is the remainder.
p(-1)=( -1)3 + 3(-1)2 - 2(-1) +7
= -1 +3 +2 +7
= 11
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