# Thread: The remainder and factor theorems 2 (what did I do wrong?)

1. ## The remainder and factor theorems 2 (what did I do wrong?)

The remainder when x^3 + ax^2 + bx + 1 is divided by (x - 5) is -14. When the cubic polynomial is divided by (x + 1), the remainder is -2. Find a and b.

To solve a and b, I did the following:

P(5) = 5^3 + a(5)^2 + b(5) + 1
= 125 + 25a + 5b + 1
= 25a + 5b + 126

25a + 5b + 126 = -14

P(-1) = (-1)^3 + a(-1)^2 + b(-1) + 1
= -1 + a - b + 1
= a - b

a - b = -2

Using the method of elimination:

25a + 5b = -140
25a - 25b = -50
30b = -190
b= -19/3

Substituting the value of b into: a - b = -2
a - (-19/3) = -2
a + 19/3 = -2
a = -25/3

But the answers are supposed to be:

a = -5, b= -3

What did I do wrong? Please show me complete working out. All help will be appreciated.

2. Originally Posted by Mr Rayon
The remainder when x^3 + ax^2 + bx + 1 is divided by (x - 5) is -14. When the cubic polynomial is divided by (x + 1), the remainder is -2. Find a and b.

To solve a and b, I did the following:

P(5) = 5^3 + a(5)^2 + b(5) + 1
= 125 + 25a + 5b + 1
= 25a + 5b + 126

25a + 5b + 126 = -14

P(-1) = (-1)^3 + a(-1)^2 + b(-1) + 1
= -1 + a - b + 1
= a - b

a - b = -2

Using the method of elimination:

25a + 5b = -140
25a - 25b = -50
30b = -190 Mr F says: -140 - (-50) = -90. So you should have 30b = -90.

b= -19/3

Substituting the value of b into: a - b = -2
a - (-19/3) = -2
a + 19/3 = -2
a = -25/3

But the answers are supposed to be:

a = -5, b= -3

What did I do wrong? Please show me complete working out. All help will be appreciated.
Note: Do yourself a favor and simplify your equations before trying to solve them. By dividing equation 1 through by 5 you get 5a + b = -28. Much easier to work with.

3. Oh wait I think I understand now! From above I was assuming that they were simulataneous equations. So to solve a and b I just need to use the quadaratic formulat on both separately?

4. Originally Posted by Mr Rayon
Oh wait I think I understand now! From above I was assuming that they were simulataneous equations. So to solve a and b I just need to use the quadaratic formulat on both separately?
No. Read my post again - the red stuff (I made an edit to make my point more obvious).

5. Originally Posted by mr fantastic
Note: Do yourself a favor and simplify your equations before trying to solve them. By dividing equation 1 through by 5 you get 5a + b = -28. Much easier to work with.
Oh...yes I understand completely now!