Given that (x+1) is a factor of p(x)=x^3+ax^2-6x+5, find the value of a
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Given that (x+1) is a factor of p(x)=x^3+ax^2-6x+5, find the value of a

[From: ] [author: ] [Date: 11-11-04] [Hit: ]
-Because x+1 is a factor then one can have x=-1. Once you have this let p(x)=0 and substitute x=-1 into p(x). You will find that it becomes (-1)^3 +a(-1)^2 - 6(-1) + 5 = 0, from this you can easily solve for a. Have fun.So,......
please help, 10 points

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x + 1 is a factor if and only if -1 is a root---i.e. p(-1) = 0. This gives an equation for a.

0 = p(-1) = (-1)^3 + a(-1)² - 6(-1) + 5 ==>

0 = a +10 ==> a = - 10.

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Because x+1 is a factor then one can have x=-1. Once you have this let p(x)=0 and substitute x=-1 into p(x). You will find that it becomes (-1)^3 +a(-1)^2 - 6(-1) + 5 = 0, from this you can easily solve for a. Have fun.

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(x + 1)*(x^2 + c*x + 5) = x^3+ax^2-6x+5
x^3 + (c+ 1)*x^2 + (c + 5)*x + 5 = x^3 + ax^2 - 6x + 5

As you can see:
c + 1 = a
c + 5 = -6

So, c = -11
a = 1 - 11 = -10
1
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