Find the fifth degree polynomial which leaves remainder 1 when divided by (x-1)^{3 and remainder -1} when divided (x+1)³

To find a fifth-degree polynomial that meets the specified conditions, we can use the Remainder Theorem and polynomial interpolation. The Remainder Theorem states that if a polynomial \( P(x) \) is divided by \( (x - c) \), the remainder is \( P(c) \). In this case, we need to construct a polynomial \( P(x) \) such that:

• When divided by \( (x - 1)^3 \), the remainder is 1.
• When divided by \( (x + 1)^3 \), the remainder is -1.

Since we are looking for a fifth-degree polynomial, we can express \( P(x) \) in the following form:

P(x) = (x - 1)^3 Q(x) + R_1(x)

Here, \( Q(x) \) is a polynomial of degree 2 (since the total degree must be 5), and \( R_1(x) \) is the remainder when divided by \( (x - 1)^3 \). Given that the remainder is 1, we can set:

R_1(x) = 1

Now, we can express \( P(x) \) as:

P(x) = (x - 1)^3 Q(x) + 1

Next, we need to satisfy the second condition regarding division by \( (x + 1)^3 \). We can write:

P(x) = (x + 1)^3 R(x) + R_2(x)

Where \( R_2(x) \) is the remainder when divided by \( (x + 1)^3 \). According to the problem, this remainder should equal -1, so:

R_2(x) = -1

Now, we can equate the two forms of \( P(x) \):

(x - 1)^3 Q(x) + 1 = (x + 1)^3 R(x) - 1

Rearranging gives us:

(x - 1)^3 Q(x) - (x + 1)^3 R(x) = -2

To solve for \( Q(x) \) and \( R(x) \), we can choose specific values for \( x \) to simplify our calculations. Let's evaluate at \( x = 1 \) and \( x = -1 \):

1. For \( x = 1 \):

(1 - 1)^3 Q(1) - (1 + 1)^3 R(1) = -2

This simplifies to:

-8R(1) = -2

Thus, \( R(1) = \frac{1}{4} \).

2. For \( x = -1 \):

(-1 - 1)^3 Q(-1) - (-1 + 1)^3 R(-1) = -2

This simplifies to:

-8Q(-1) = -2

Thus, \( Q(-1) = \frac{1}{4} \).

Now, we can assume a simple form for \( Q(x) \) and \( R(x) \). Let's try:

Q(x) = ax^2 + bx + c

R(x) = dx^2 + ex + f

Since we have two equations and two unknowns, we can solve for \( a, b, c, d, e, \) and \( f \) using the conditions we derived. However, for simplicity, we can also directly construct \( P(x) \) based on our earlier findings. We can express \( P(x) \) as:

P(x) = A(x - 1)^3 + 1

And we can find \( A \) such that:

P(-1) = -1

Substituting \( x = -1 \):

P(-1) = A(-2)^3 + 1 = -1

This gives us:

-8A + 1 = -1

Thus, \( A = \frac{1}{4} \).

Finally, substituting \( A \) back into our polynomial gives:

P(x) = \frac{1}{4}(x - 1)^3 + 1

To express this as a fifth-degree polynomial, we can expand it:

P(x) = \frac{1}{4}(x^3 - 3x^2 + 3x - 1) + 1

P(x) = \frac{1}{4}x^3 - \frac{3}{4}x^2 + \frac{3}{4}x + \frac{3}{4}

However, we need to ensure that the degree is 5. Therefore, we can add a term of degree 5, say \( kx^5 \), where \( k \) is a constant. Thus, the final polynomial can be expressed as:

P(x) = kx^5 + \frac{1}{4}x^3 - \frac{3}{4}x^2 + \frac{3}{4}x + \frac{3}{4}

By adjusting \( k \), we can ensure that the polynomial remains a fifth-degree polynomial while satisfying the remainder conditions. This gives us the flexibility to create a family of solutions based on the value of \( k \). For instance, if we set \( k = 0 \), we revert to a cubic polynomial, but any non-zero \( k \) will yield a valid fifth-degree polynomial.