0.36x^2 + 0.63x + 1.17 The given expression can be rewritten as \(\mathrm{a}(4\mathrm{x}^2 + 7\mathrm{x} + 13)\), where a is...

1. TRANSLATE the problem information

• Given information:
  • Original expression: \(0.36\mathrm{x}^2 + 0.63\mathrm{x} + 1.17\)
  • Factored form: \(\mathrm{a}(4\mathrm{x}^2 + 7\mathrm{x} + 13)\), where a is unknown
  • These two expressions are equivalent

2. INFER the approach

• Since the expressions are equivalent, I can set them equal to each other
• This will allow me to find the value of a by comparing coefficients

3. Set up the equation

\(0.36\mathrm{x}^2 + 0.63\mathrm{x} + 1.17 = \mathrm{a}(4\mathrm{x}^2 + 7\mathrm{x} + 13)\)

4. SIMPLIFY by expanding the right side

Using the distributive property:

\(\mathrm{a}(4\mathrm{x}^2 + 7\mathrm{x} + 13) = 4\mathrm{ax}^2 + 7\mathrm{ax} + 13\mathrm{a}\)

So: \(0.36\mathrm{x}^2 + 0.63\mathrm{x} + 1.17 = 4\mathrm{ax}^2 + 7\mathrm{ax} + 13\mathrm{a}\)

5. INFER the coefficient relationships

• For equivalent polynomials, coefficients of like terms must be equal
• This gives me three equations:
  • x² coefficients: \(0.36 = 4\mathrm{a}\)
  • x coefficients: \(0.63 = 7\mathrm{a}\)
  • constant terms: \(1.17 = 13\mathrm{a}\)

6. SIMPLIFY to solve for a

Using the first equation: \(0.36 = 4\mathrm{a}\)

Divide both sides by 4: \(\mathrm{a} = 0.36 \div 4 = 0.09\)

Answer: 0.09 (or \(\frac{9}{100}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may misunderstand what "can be rewritten as" means and fail to set up the equivalence equation properly.

Instead of setting the expressions equal, they might try to substitute values or attempt to factor the original expression directly without using the given factored form. This leads to confusion about how to proceed and often results in guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up the coefficient equations but make arithmetic errors when dividing.

For example, calculating \(0.36 \div 4\) incorrectly as 0.9 instead of 0.09, or mixing up which coefficient goes with which equation. This may lead them to select an incorrect decimal value that's off by a factor of 10.

The Bottom Line:

This problem tests whether students can bridge the gap between algebraic equivalence and coefficient comparison. The key insight is recognizing that "can be rewritten as" means the expressions are identical, so their coefficients must match perfectly.