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.