when ax^2+bx+c is divided by x-1, the remainder is 10 if the same poly. is divided by x+2 the remainder is 1. x+1 is a factor of the polynomial. Find a,b, and c.
let f(x) = ax^2 + bx + c
given f(1) = 10
10 = a + b + c (#1)
given f(-2) = 1
1 = 4a - 2b + c (#2)
given f(-1) = 0 , .... (it said x+1 is a factor)
0 = a - b + c (#3)
#1 - #3
10 = 0 + 2b + 0
b = 5
#2 - #3
1 = 3a - b + 0 , but b=5
6 = 3a
a = 2
in #1
2 + 5 + c = 10
c = 3
a=2, b=5, c=3
To find the values of a, b, and c, we can use the remainder theorem and the fact that x + 1 is a factor of the polynomial.
1. When a polynomial is divided by (x - p), the remainder is given by substituting p into the polynomial.
2. Given that the remainder when dividing ax^2 + bx + c by x - 1 is 10, we substitute x = 1 into the polynomial to get a(1)^2 + b(1) + c = 10, which simplifies to a + b + c = 10.
3. Similarly, when dividing ax^2 + bx + c by x + 2, the remainder is 1. Substituting x = -2 into the polynomial gives a(-2)^2 + b(-2) + c = 1, which simplifies to 4a - 2b + c = 1.
4. Finally, since x + 1 is a factor, substituting x = -1 into the polynomial yields a(-1)^2 + b(-1) + c = 0, which simplifies to a - b + c = 0.
Now, we have the following system of equations:
a + b + c = 10 (Equation 1)
4a - 2b + c = 1 (Equation 2)
a - b + c = 0 (Equation 3)
Solving this system of equations will give us the values of a, b, and c.
One way to solve this system is by elimination:
- Add Equation 2 and Equation 3 together to eliminate "b" and simplify the system.
- Subtract Equation 3 from Equation 1 to eliminate "c" and simplify the system.
- Solve the resulting system of two equations with two variables to find the values of a, b, and c.
Let's solve it step-by-step:
1. Add Equation 2 and Equation 3:
(4a - 2b + c) + (a - b + c) = 1 + 0
Simplifying, we get:
5a - 3b + 2c = 1 (Equation 4)
2. Subtract Equation 3 from Equation 1:
(a + b + c) - (a - b + c) = 10 - 0
Simplifying, we get:
2b = 10
b = 5
3. Substitute b = 5 into Equation 4:
5a - 3(5) + 2c = 1
Simplifying further:
5a - 15 + 2c = 1
5a + 2c = 16 (Equation 5)
4. Substitute b = 5 into Equation 3:
a - 5 + c = 0
a + c = 5 (Equation 6)
5. Solve the system of equations formed by Equation 5 and Equation 6.
Subtract Equation 5 from Equation 6:
(a + c) - (5a + 2c) = 5 - 16
Simplifying, we get:
-4a - c = -11
6. Now, we have a new equation. Combine it with Equation 5:
-4a - c = -11 (Equation 7)
5a + 2c = 16 (Equation 5)
7. Multiply Equation 5 by 2 and Equation 7 by 5 to get a new system of equations:
2(5a + 2c) = 2(16) --> 10a + 4c = 32 (Equation 8)
5(-4a - c) = 5(-11) --> -20a - 5c = -55 (Equation 9)
8. Add Equation 8 and Equation 9:
(10a + 4c) + (-20a - 5c) = 32 - 55
Simplifying, we get:
-10a - c = -23
9. Now, we have a new equation. Combine it with Equation 6:
-10a - c = -23 (Equation 10)
a + c = 5 (Equation 6)
10. Multiply Equation 6 by 1 and Equation 10 by -10 to get a new system of equations:
(a + c) + (-10a - c) = 5 - 23
Simplifying, we get:
-9a = -18
11. Divide both sides of Equation 10 by -9:
a = 2
12. Substitute a = 2 into Equation 6 to find c:
2 + c = 5
c = 3
Now we have found the values of a, b, and c:
a = 2
b = 5
c = 3