The polynomial P(x)=ax^3+bx^2+cx+d leaves a remainder of 10x+11 when it is divided by x^2−1. It leaves a remainder of −6 when it is divided by x^2+x+1. What is the value of bc−ad
To find the value of bc−ad, we need to solve a system of equations using the remainder theorem.
The remainder theorem states that when a polynomial P(x) is divided by (x - r), the remainder is P(r). Using this theorem, we can set up two equations to find the remainders when P(x) is divided by x^2−1 and x^2+x+1, respectively.
Equation 1: P(x) divided by (x^2−1) leaves a remainder of 10x+11
Equation 2: P(x) divided by (x^2+x+1) leaves a remainder of -6
Step 1: For Equation 1, set up synthetic division with divisor (x^2−1) and remainder 10x+11. The dividend will have the coefficients of a, b, c, d: (a, b, c, d).
The synthetic division will look like this:
_________
x^2-1 | a b c d
__________________
Performing the synthetic division, we get:
x^2-1 | a b c d
__________________
Step 2: For Equation 2, set up synthetic division with divisor (x^2+x+1) and remainder -6. The dividend will still have the coefficients of a, b, c, d: (a, b, c, d).
The synthetic division will look like this:
___________
x^2+x+1 | a b c d
___________________
Performing the synthetic division, we get:
x^2+x+1 | a b c d
___________________
Step 3: Now, equate the remainders in each synthetic division to the given remainders.
From Equation 1: Remainder = 10x+11
From Equation 2: Remainder = -6
So, we have the following equations:
Equation 1: Remainder = 10x+11
Equation 2: Remainder = -6
Step 4: Equate the expressions for the remainders to find the values of a, b, c, and d.
From Equation 1:
10x + 11 = (x - 1)(x^2 - 1)
Expanding the right side, we get:
10x + 11 = (x - 1)(x + 1)(x - 1)
Simplifying further, we have:
10x + 11 = (x^2 - 1)(x - 1)
10x + 11 = (x^3 - x - x + 1)(x - 1)
10x + 11 = (x^3 - 2x + 1)(x - 1)
10x + 11 = x^4 - 3x^2 + 3x - x^3 + 2x - 1
10x + 11 = x^4 - x^3 - 3x^2 + 5x - 1
Comparing the coefficients of x on both sides, we get:
10 = 5 → 5a - b = 10
11 = -1 → -c + 2d = 11
From Equation 2:
-6 = (x^2 + x + 1)(x^2 + x + 1)
Expanding the right side, we get:
-6 = (x^2 + x + 1)^2
-6 = (x^2 + x + 1)(x^2 + x + 1)
-6 = x^4 + 2x^3 + 3x^2 + 2x + 1
Comparing the coefficients of x on both sides, we get:
0 = 2 → 2a = 0
0 = 1 → b + 2c = 0
-6 = 1 → d = -6
From the equations we found:
5a - b = 10
-c + 2d = 11
2a = 0
b + 2c = 0
d = -6
We can solve this system of equations to find the values of a, b, c, and d.
Solving the equations, we get:
a = 0
b = 0
c = 0
d = -6
Step 5: Substitute the values of a, b, c, and d into the expression for bc−ad.
bc − ad = 0 * 0 - 0 * (-6)
= 0
Therefore, the value of bc−ad is 0.