Answered

Match the equivalent expressions. Justify your answers.


  1. $\left(x^2-x-8\right) \div (x-4)$
  2. $\left(x^2-x+8\right) \div (x-4)$
  3. $\left(x^2+x-8\right) \div (x-4)$
  4. $\left(x^2+x+8\right) \div (x-4)$


A$x+3+\frac{4}{x-4}$
B$x+5+\frac{12}{x-4}$
C$x+5+\frac{28}{x-4}$
D$x+3+\frac{20}{x-4}$

Verified Answer :

  • Analyze each expression using polynomial division with the divisor \(x-4\).
  • Match 13 to A: \((x^2-x-8)/(x-4) = x+3+4/(x-4)\)
  • Match 14 to D: \((x^2-x+8)/(x-4) = x+3+20/(x-4)\)
  • Match 15 to B: \((x^2+x-8)/(x-4) = x+5+12/(x-4)\)
  • Match 16 to C: \((x^2+x+8)/(x-4) = x+5+28/(x-4)\)
  • Final Matches: 13-A, 14-D, 15-B, 16-C.

Explanation

  1. Analysis and Methodology

To express each quadratic expression in the form \((x+k) + \frac{r}{x-4}\), we perform polynomial long division by the divisor \((x-4)\). For a general quadratic of the form \(x^2 + bx + c\), we can write:
\[\frac{x^2 + bx + c}{x-4} = \frac{x(x-4) + 4x + bx + c}{x-4} = x + \frac{(4+b)x + c}{x-4} = x + (4+b) + \frac{4(4+b) + c}{x-4}.\]
Thus, the quotient part is \(x + (4+b)\) and the remainder \(r\) is \(16 + 4b + c\). We will apply this to each expression.

  1. Solving 13 and 14
  2. For \((x^2 - x - 8) \div (x - 4)\):

Here \(b = -1\) and \(c = -8\).
The quotient constant is \(4 + (-1) = 3\).
The remainder is \(16 + 4(-1) - 8 = 16 - 4 - 8 = 4\).
Thus, \((x^2 - x - 8) \div (x - 4) = x + 3 + \frac{4}{x-4}\). This matches option A.

  1. For \((x^2 - x + 8) \div (x - 4)\):

Here \(b = -1\) and \(c = 8\).
The quotient constant is \(4 + (-1) = 3\).
The remainder is \(16 + 4(-1) + 8 = 16 - 4 + 8 = 20\).
Thus, \((x^2 - x + 8) \div (x - 4) = x + 3 + \frac{20}{x-4}\). This matches option D.

  1. Solving 15 and 16
  2. For \((x^2 + x - 8) \div (x - 4)\):

Here \(b = 1\) and \(c = -8\).
The quotient constant is \(4 + 1 = 5\).
The remainder is \(16 + 4(1) - 8 = 16 + 4 - 8 = 12\).
Thus, \((x^2 + x - 8) \div (x - 4) = x + 5 + \frac{12}{x-4}\). This matches option B.

  1. For \((x^2 + x + 8) \div (x - 4)\):

Here \(b = 1\) and \(c = 8\).
The quotient constant is \(4 + 1 = 5\).
The remainder is \(16 + 4(1) + 8 = 16 + 4 + 8 = 28\).
Thus, \((x^2 + x + 8) \div (x - 4) = x + 5 + \frac{28}{x-4}\). This matches option C.

Examples

Polynomial division is fundamental in algebra for simplifying rational functions, which appear frequently in fields like engineering and economics. For instance, when analyzing the cost efficiency of a manufacturing process where the cost per unit is modeled by a rational function \(C(x) = \frac{ax^2+bx+c}{x-k}\), engineers use polynomial division to find the "fixed cost" component (the quotient) and the "variable factor" (the remainder term). This helps in understanding how costs stabilize as the number of units \(x\) increases.

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