Remainder Theorem Question:
For what value of b will the polynomial P(x) = 4x^3-3x^2+bx+6 have the same remainder when it is divided by x-1 and by x+3
Do the polynomial long divisions for each divisor. Then set the two remainders equal to each other. Both will contain a "b" term. Solve for b.
To find the value of b that will make the polynomial P(x) have the same remainder when divided by x-1 and x+3, we can use the Remainder Theorem.
The Remainder Theorem states that if a polynomial P(x) is divided by a linear divisor of the form x-a, the remainder is equal to P(a).
So, let's find the remainders when P(x) is divided by x-1 and x+3.
When P(x) is divided by x-1, the remainder is P(1). We can substitute x=1 into the polynomial:
P(1) = 4(1)^3 - 3(1)^2 + b(1) + 6
= 4 - 3 + b + 6
= b + 7
When P(x) is divided by x+3, the remainder is P(-3). Substituting x=-3 into the polynomial:
P(-3) = 4(-3)^3 - 3(-3)^2 + b(-3) + 6
= -108 - 27 - 3b + 6
= -3b - 129
Since we want the polynomial to have the same remainder when divided by both x-1 and x+3, we can set the remainders equal to each other:
b + 7 = -3b - 129
Now, we can solve this equation for b:
4b = -136
b = -34
Therefore, the value of b that will make the polynomial P(x) have the same remainder when divided by x-1 and x+3 is -34.
To find the value of b that will make the polynomial P(x) have the same remainder when divided by x-1 and x+3, we can apply the Remainder Theorem.
According to the Remainder Theorem, if a polynomial P(x) is divided by x - c, the remainder is equal to P(c). Therefore, we need to set up two equations using the Remainder Theorem.
First, let's consider the division of P(x) by x-1. The remainder will be equal to P(1):
Remainder when dividing by x-1 = P(1) = 4(1)^3 - 3(1)^2 + b(1) + 6
Next, let's consider the division of P(x) by x+3. The remainder will be equal to P(-3):
Remainder when dividing by x+3 = P(-3) = 4(-3)^3 - 3(-3)^2 + b(-3) + 6
We want the remainder to be the same for both divisions. Setting the two expressions equal to each other, we have:
4(1)^3 - 3(1)^2 + b(1) + 6 = 4(-3)^3 - 3(-3)^2 + b(-3) + 6
Simplifying the equation, we get:
4 - 3 + b + 6 = -108 - 27b + 6
Combine like terms:
b + 7 = -102 - 27b
Bring the terms involving b to one side:
b + 27b = -102 - 7
Combine like terms:
28b = -109
Finally, divide both sides of the equation by 28:
b = -109/28
Therefore, the value of b that will make the polynomial have the same remainder when divided by x-1 and x+3 is -109/28.