Problem 41 - Precalculus with Calculus Previews

Question

In Problems 41 and \(42,\) use long division to find a value of \(k\) such that \(f(x)\) is divisible by \(d(x)\). $$ f(x)=x^{4}+x^{3}+3 x^{2}+k x-4 ; d(x)=x^{2}-1 $$

Step-by-Step Solution

Answer

The value of \( k \) is \(-1\).

Step 1: Set Up the Long Division

We need to divide the polynomial \( f(x) = x^4 + x^3 + 3x^2 + kx - 4 \) by \( d(x) = x^2 - 1 \). Set up the long division, aligning powers of \( x \) beneath each other.

Step 2: Divide the Leading Terms

Divide the leading term of \( f(x) \), which is \( x^4 \), by the leading term of \( d(x) \), which is \( x^2 \). This gives \( x^2 \). Place \( x^2 \) on top of the division.

Step 3: Multiply and Subtract

Multiply \( x^2 \) by \( d(x) = x^2 - 1 \), resulting in \( x^4 - x^2 \). Subtract this from \( f(x) \): \[ (x^4 + x^3 + 3x^2 + kx - 4) - (x^4 - x^2) = x^3 + 4x^2 + kx - 4 \].

Step 4: Repeat Division

Divide the new leading term, \( x^3 \), by \( x^2 \). This gives \( x \). Place \( x \) on top next to \( x^2 \).

Step 5: Multiply and Subtract Again

Multiply \( x \) by \( d(x) = x^2 - 1 \), resulting in \( x^3 - x \). Subtract this from the previous result: \[ (x^3 + 4x^2 + kx - 4) - (x^3 - x) = 4x^2 + (k+1)x - 4 \].

Step 6: Final Division and Subtraction

Divide \( 4x^2 \) by \( x^2 \), resulting in \( 4 \). Multiply \( 4 \) by \( d(x) = x^2 - 1 \), resulting in \( 4x^2 - 4 \). Subtract to find the remainder: \[ (4x^2 + (k+1)x - 4) - (4x^2 - 4) = (k+1)x \].

Step 7: Set Remainder to Zero and Solve for k

Since \( f(x) \) must be divisible by \( d(x) \), the remainder \((k+1)x\) must equal zero for all \( x \). Therefore, \( k+1 = 0 \). Solve for \( k \) to find \( k = -1 \).

Key Concepts

Divisible Polynomials

To determine if a polynomial is divisible by another, we need to check if the remainder of the division is zero. When we say a polynomial \( f(x) \) is divisible by \( d(x) \), we're implying that there's a factor in common, and the division results in a whole polynomial without any leftovers. Long division is a great tool to verify this. It helps break down complex polynomial expressions.

• Set up the division in the same way you would with numbers, ensuring that like terms (the same power of \( x \)) are lined up.
• After performing the division, calculate the remainder at the end.
• If the remainder is zero, we say that \( f(x) \) is divisible by \( d(x) \).

Understanding divisible polynomials is essential because many problems in algebra require polynomial factorization or simplification, and ensuring divisibility allows you to break complex problems into more manageable parts. Remember, a remainder of zero confirms that the polynomial has been fully divided.

Polynomial Division

Polynomial division can initially seem daunting, but it's similar to dividing numbers. Long division is used to divide one polynomial by another. In our example, we divided \( f(x) = x^4 + x^3 + 3x^2 + kx - 4 \) by \( d(x) = x^2 - 1 \). Here’s how you tackle it:

• Begin by dividing the leading term of the dividend (the higher degree polynomial) by the leading term of the divisor (the lower degree polynomial).
• Write the result as a new term in the quotient.
• Multiply the entire divisor by this new term and subtract the result from the current dividend, just as you would with numbers.
• Repeat the process with the new polynomial created after subtraction, until the degree of the new polynomial is less than the degree of the divisor.

This technique systematically narrows down the quotient and remainder, allowing you to see if the divisor evenly fits into the dividend.

Finding Polynomial Remainders

The remainder in a polynomial division tells us what part of the dividend has not been perfectly divided by the divisor. In the exercise, after performing the steps of long division, the remainder was found to be \((k+1)x\). For \( f(x) \) to be divisible by \( d(x) \), this remainder must be zero.

• Set the remainder to zero, as this indicates complete divisibility.
• Solve the resulting equation for the unknown constant or term. In this exercise, \( k+1 = 0 \) led to \( k = -1 \).
• Once solved, it verifies that the polynomial division resulted in a zero remainder, confirming divisibility.

By understanding how to find and use remainders in polynomial division, you're able to not just divide polynomials but also adjust values within them to achieve divisibility. This skill is crucial in algebra for simplifying and solving polynomial equations effectively.