The expression ax^2 + bx + c is divisible exactly by (x - 1), has a remainder of 2 when it is divided by (x + 1), and has a remainder of 6 when it is divided by (x - 2). Find a, b and c?
using the remainder theorem, f(a) is the remainder when dividing by (x-a), we have
f(1) = 0
f(-1) = 2
f(2) = 6
So, now you have three equations to use:
a+b+c = 0
a-b+c = 2
4a+2b+c = 6
7/3 x^2 - x - 4/3
check it to be sure.
thx
I didn't understand the last part
To find the values of a, b, and c, we can use the concept of polynomial division and the remainder theorem.
Let's start by using the information given to solve for a, b, and c.
We know that the expression ax^2 + bx + c is divisible exactly by (x - 1), which means that when we divide ax^2 + bx + c by (x - 1), the remainder is zero. This can be expressed as:
(ax^2 + bx + c) ÷ (x - 1) = 0
We can perform polynomial long division to find the quotient and remainder:
a(x - 1) + (a + b)(x - 1) + c = 0
=> a(x - 1) + ax - a + bx - b + c = 0
=> ax - a + bx - b + a(x - 1) + c = 0
=> ax - a + bx - b + ax - a + c = 0
=> 2ax - 2a + bx - b + c = 0
=> (2a + b)x - (2a + b) + c = 0
Since this expression is divisible exactly by (x - 1), the coefficient of x in the resulting expression should be zero. Therefore, we have:
2a + b = 0 (Equation 1)
-(2a + b) + c = 0 (Equation 2)
Now, we also know that the expression has a remainder of 2 when divided by (x + 1). This can be expressed as:
(ax^2 + bx + c) ÷ (x + 1) = R
Using polynomial long division again, we have:
ax^2 + bx + c
x + 1 | ax^2 + (a + b)x + c
- (ax^2 + ax)
________________
(a + b - ax) + c
Since the remainder is 2, we have:
(a + b - ax) + c = 2 (Equation 3)
Similarly, we know that the expression has a remainder of 6 when divided by (x - 2). This can be expressed as:
(ax^2 + bx + c) ÷ (x - 2) = R
Using polynomial long division, we have:
ax^2 + bx + c
x - 2 | ax^2 + (a + b)x + c
- (ax^2 - 2ax)
________________
(a + b + 2ax) + c
Since the remainder is 6, we have:
(a + b + 2ax) + c = 6 (Equation 4)
Now we have a system of linear equations (Equations 1, 2, 3, and 4) with three variables, a, b, and c. We can solve this system to find the values of a, b, and c.
By solving these equations simultaneously, we can find the values of a, b, and c:
From Equation 1:
2a + b = 0 ==> b = -2a
Substituting b = -2a into Equation 2:
-(2a + (-2a)) + c = 0
-4a + c = 0 ==> c = 4a
Substituting b = -2a and c = 4a into Equation 4:
(a + (-2a) + 2a x) + 4a = 6
4a + 2ax + 4a = 6
8a + 2ax = 6
We can simplify this equation by dividing by 2:
4a + ax = 3
Now, we can solve for a by equating the coefficients of x on both sides:
a = 3 (Multiply by 2 and divide by (2 + x))
Substituting the value of a back into b = -2a:
b = -2(3)
b = -6
And substituting the value of a back into c = 4a:
c = 4(3)
c = 12
Therefore, the values of a, b, and c are a = 3, b = -6, and c = 12, respectively.