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## #1 2017-06-07 22:03:39

KayleXYZ
Member
Registered: 2017-06-07
Posts: 1

### Encountered some problem in this question....

Hi! i used your website to refresh my math skill and it's good so far until i encountered an error in this question about remainder and factor theorem here's the link: http://www.mathopolis.com/questions/q.php?id=231&site=1&ref=/algebra/polynomials-remainder-factor.html&qs=0

i tried to solve the problem and come up with the answer a = 1, b = -4, c = -4 but the answer is not there and i tried to pick a random answer so i can see the correct answer and the correct answer is b, but when i substitute the value to the given equation the result is not equals to 0 here's how i checked it :

(-1^3) + (-1)(-1^2) + (-4)(-1) + (4) = 0

-1 - 1 + 4 + 4 = 0

-2 + 4 + 4 = 0

2 + 4 = 0

6 = 0

Here's the result when i substitute my answer:

(-1^3) + (1)(-1^2) + (-4)(-1) + (-4) = 0

-1 + 1 + 4 - 4 = 0

0 = 0

Please correct me if i'm wrong.

Here's my solution:

-1 + a + b + c

a + b + c = 1

8 + 4a + 2b + c

4a + 2b + c = -8

-8 + 4a -2b + c

4a - 2b + c = 8

a + b + c - 4a - 2b - c = 1 + 8

-3a - b = 9

a + b + c - 4a + 2b - c = 1 - 8

-3a + 3b = -7

-3a - b + 3a - 3b = 9 + 7

-4b = 16

b = -4

-1 + a + 4 + c = 0

3 + a + c = 0

a + c = -3

8 + 4a - 8 + c = 0

4a + c = 0

-8 + 4a + 8 + c = 0

4a + c = 0

a + c -4a - c = -3 - 0

-3a = -3

a = 1

-1 + 1 + 4 + c = 0

4 + c = 0

c = -4

check:

-1 + 1 + 4 - 4 = 0

0 = 0

a = 1, b = -4, c = -4

Last edited by KayleXYZ (2017-06-07 22:51:12)

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## #2 2017-06-10 21:32:19

bob bundy
Registered: 2010-06-20
Posts: 8,371

### Re: Encountered some problem in this question....

hi  KayleXYZ

I didn't solve this that way.  I just expanded (x-1)(x+2)(x-2) and looked at the coefficients.

But using your approach the three equations should be

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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