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Old 28th April 2012, 06:56 PM   #1 (permalink)
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Factor and Remainder Theorem

Hi, I have two questions that I can't solve:

Q1) Given that X - P is a factor of the expression X^2 - (P + 2)X - P^2 + 4P + 8, calculate the possible values of P
A1) -2, 4
My Attempt:
Let X^2 - (P+2)X - P^2 + 4P + 8 by f(x)
f(P) = 0
(P)^2 - (P+2)(P) - P^2 + 4P + 8 = 0
P^2 - P^2 - 2P - P^2 + 4P + 8 = 0
-2P + 4P + 8 = 0
2P = -8
P = -4

I have no ideal why I keep getting -4 instead of two numbers

Q2) A cubic equation f(x) = 0 has roots -3, -1 and 4. f(x) leaves a remainder of 12 when it is divided by x+2. Find the remainder when f(x) is divided by x^2-2x-1.
A2) -16x - 20

My Attempt:
(X+3)(X+1)(X-4) = 12
(X+3)(X^2-3X-4) = 12
X^3-3X^2-4X+3X^2-9X-12 = 12
X^3 -13X-12 = 12
X^3-13X-24 = 0
Division:
X^3-13X-24 = 0 / x^2-2x-1 = -8x-22

Any help is appreciate, thanks

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Old 28th April 2012, 07:30 PM   #2 (permalink)
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Re: Factor and Remainder Theorem

Originally Posted by danny2010 View Post
Hi, I have two questions that I can't solve:

Q1) Given that X - P is a factor of the expression X^2 - (P + 2)X - P^2 + 4P + 8, calculate the possible values of P
A1) -2, 4
My Attempt:
Let X^2 - (P+2)X - P^2 + 4P + 8 by f(x)
f(P) = 0
(P)^2 - (P+2)(P) - P^2 + 4P + 8 = 0
P^2 - P^2 - 2P - P^2 + 4P + 8 = 0
-2P + 4P + 8 = 0
2P = -8
P = -4

I have no ideal why I keep getting -4 instead of two numbers
Where the blue -P^2 go? Lol.

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Old 28th April 2012, 07:41 PM   #3 (permalink)
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Re: Factor and Remainder Theorem

Q2) A cubic equation f(x) = 0 has roots -3, -1 and 4. f(x) leaves a remainder of 12 when it is divided by x+2. Find the remainder when f(x) is divided by x^2-2x-1.
A2) -16x - 20


a(-2+3)(-2+1)(-2-4) = 12
a(1)(-1)(-6)=12
a=2

Thus f(x)= 2(X+3)(X+1)(X-4)

Continue to divide and find the remainder.

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Last edited by aforl; 28th April 2012 at 07:41 PM.
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Old 1st May 2012, 10:58 PM   #4 (permalink)
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Re: Factor and Remainder Theorem

Originally Posted by aforl View Post
Where the blue -P^2 go? Lol.
I just subtract them together, P^2-P^2-P^2 = 0 since they are same term but I just realize that if I neutralize that first two P^2, I can shift that blue -P^2 over to the right hand side then the rest of the left hand side, I shift it to right hand side and factorize them, the answer is P=4 and P=-2, it is pretty weird. Thanks for your help!

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Old 2nd May 2012, 08:45 PM   #5 (permalink)
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Re: Factor and Remainder Theorem

Originally Posted by danny2010 View Post
I just subtract them together, P^2-P^2-P^2 = 0 since they are same term but I just realize that if I neutralize that first two P^2, I can shift that blue -P^2 over to the right hand side then the rest of the left hand side, I shift it to right hand side and factorize them, the answer is P=4 and P=-2, it is pretty weird. Thanks for your help!
Are you still thinking that P^2-P^2-P^2 = 0 ?!

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