Problem 46

Question

\(39-51\) . Use synthetic division and the Remainder Theorem to evaluate \(P(c) .\) $$ P(x)=6 x^{5}+10 x^{3}+x+1, \quad c=-2 $$

Step-by-Step Solution

Verified
Answer
The value of \( P(-2) \) is \(-273\).
1Step 1: Organize Coefficients
Write down the coefficients of the polynomial \( P(x) = 6x^5 + 0x^4 + 10x^3 + 0x^2 + x + 1 \). These are \([6, 0, 10, 0, 1, 1]\). Note the coefficients for terms with zero coefficients are included.
2Step 2: Set Up Synthetic Division
To use synthetic division with \(c = -2\), write \(-2\) outside a division-like symbol and place the coefficients \([6, 0, 10, 0, 1, 1]\) inside.
3Step 3: Perform Synthetic Division
Start the synthetic division by bringing down the first coefficient, which is 6. Multiply this 6 by \(-2\) and write the result under the next coefficient. Add this result to the coefficient above it. Continue this process:1. Bring down the 6.2. \( -2 \times 6 = -12 \) (write under 0).3. \( 0 + (-12) = -12 \).4. \( -2 \times -12 = 24 \) (write under 10).5. \( 10 + 24 = 34 \).6. \( -2 \times 34 = -68 \) (write under 0).7. \( 0 + (-68) = -68 \).8. \( -2 \times -68 = 136 \) (write under 1).9. \( 1 + 136 = 137 \).10. \( -2 \times 137 = -274 \) (write under last 1).11. \( 1 - 274 = -273 \).The result of synthetic division is: \([6, -12, 34, -68, 137, -273]\).
4Step 4: Interpret the Remainder
The remainder of the division, \(-273\), corresponds to \(P(c)\) where \(c = -2\). According to the Remainder Theorem, this is the value of \(P(-2)\).

Key Concepts

Polynomial CoefficientsRemainder TheoremPolynomial EvaluationDivision Algorithm in Algebra
Polynomial Coefficients
Polynomial coefficients are the numbers in front of the variables in a polynomial expression. For a given polynomial like \( P(x) = 6x^5 + 10x^3 + x + 1 \), each term has a coefficient which dictates the term's "weight" or influence on the polynomial. Coefficients can be zero, which means the corresponding term is absent.
  • The coefficients in our polynomial are: \([6, 0, 10, 0, 1, 1]\).
  • Note the zeroes for the \(x^4\) and \(x^2\) terms, indicating their absence.
When setting up problems, especially for synthetic division, it's crucial to include these zero coefficients to maintain the integrity of the polynomial structure.
Remainder Theorem
The Remainder Theorem is a handy tool in polynomial division. It states that the remainder of the division of a polynomial \(P(x)\) by a linear divisor \(x - c\) is equivalent to \(P(c)\), the value of the polynomial evaluated at \(c\). This means that once you perform synthetic division and obtain a remainder, you have directly found \(P(c)\).
  • In our example, dividing \(P(x)\) by \(x + 2\) yields a remainder.
  • The remainder \(-273\) is \(P(-2)\).
This theorem simplifies finding the value of a polynomial at a specific point, which is vital in simplifying algebraic computations.
Polynomial Evaluation
Polynomial evaluation involves finding the result of a polynomial expression for a specific value of the variable. When given \(P(x) = 6x^5 + 10x^3 + x + 1\) and asked to evaluate at \(x = -2\), we substitute \(-2\) for \(x\) and compute the outcome. Synthetic division aids this process by breaking down the polynomial using its coefficients.
  • The synthetic division method streamlines this evaluation.
  • Manual substitution would be tedious for higher degrees.
As seen, synthetic division not only provides the remainder swiftly but also evaluates the polynomial efficiently.
Division Algorithm in Algebra
The division algorithm in algebra refers to the procedure of dividing one polynomial by another to yield a quotient and a remainder. This is similar to long division in arithmetic. Synthetic division is a shortcut for the division algorithm specifically designed when dividing by a linear factor. The algorithm breaks down as follows:
  • Arrange the polynomial with all coefficients, including zero coefficients.
  • Use the root of the divisor \(x - c\) as the synthetic division number.
  • Consistently perform arithmetic to simplify and deduce the quotient and remainder.

In practice, synthetic division simplifies operations dramatically over the classic polynomial long division, focusing computational ease and speed.