Fraction Word Problem Solving with Answers for Grade 4

Answer: Excellent point! The main difference lies in the expression structure:

• The remarkable identity
  \((a+b)^2\)
  applies only when you have a square of a sum, meaning an expression like
  \((x+3)^2\).
• Double distribution applies to any multiplication of two binomials like
  \((x+2)(x+5)\) or \((a+b)(c+d)\).

In the case of
\((a+b)^2\),
it's actually a special case of double distribution:
\((a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\).
The remarkable identity saves you time by avoiding intermediate steps.

Example:
\((x+4)^2\) → Direct remarkable identity: \(x^2 + 8x + 16\)
But \((x+2)(x+5)\) → Double distribution: \(x^2 + 5x + 2x + 10 = x^2 + 7x + 10\)

Answer: To distinguish between the two methods, examine the expression structure:

• For the difference of two squares like \(x^2 - 16\), always look for a subtraction of two perfect square terms: \(x^2 - 4^2\). This matches the form \(a^2 - b^2\), so use the identity \(a^2 - b^2 = (a+b)(a-b)\).
• For common factor extraction, look for a factor present in all terms: for example, in \(3x^2 + 6x\), the factor \(3x\) is common.

In your example \(x^2 - 16\): \(x^2\) is the square of \(x\) and \(16\) is the square of \(4\). So it's \(x^2 - 4^2\), which factors into \((x+4)(x-4)\).

Tip: If you see a subtraction of two squared expressions, immediately think of the remarkable identity \(a^2 - b^2 = (a+b)(a-b)\).

Answer: Absolutely! Here are mnemonic tricks for each identity:

• \((a+b)^2 = a^2 + 2ab + b^2\) → "Sum squared, first square, double product, second square"
• \((a-b)^2 = a^2 - 2ab + b^2\) → "Difference squared, first square, MINUS double product, second square"
• \((a+b)(a-b) = a^2 - b^2\) → "Sum times difference, first square MINUS second square"

Another method is to remember that the first two identities have the same structure (first square, plus/minus double product, plus second square), and the third is simply the difference of squares.

Practice with simple examples like \((x+1)^2\), \((x-1)^2\), and \((x+1)(x-1)\) until the formulas become automatic!

Answer: The most reliable method to verify a factorization is to expand your result and see if you recover the original expression.

For example, if you factorized \(x^2 - 9\) into \((x+3)(x-3)\), expand \((x+3)(x-3)\):
\(x^2 - 3x + 3x - 9 = x^2 - 9\) ✓

Other tips for verification:

• Ensure all terms of the original expression appear in your factorization
• Check if you can factorize further (for example, after extracting a common factor)
• Use numerical values to test: choose a value for \(x\) (like \(x=2\)) and calculate both expressions to see if they give the same result

This numerical substitution technique is particularly useful for detecting sign errors.

Answer: Here's a systematic approach to choose the right method:

1. Recognize the structure: First look at the general form of the expression
2. For expansion:
   • A single factor multiplying parentheses → Simple distribution: \(k(a+b) = ka + kb\)
   • Two parentheses multiplying each other → Double distribution: \((a+b)(c+d)\)
   • A binomial square → Remarkable identity: \((a±b)^2\)
3. For factorization:
   • A common factor to all terms → Common factor extraction
   • A subtraction of two squares → Remarkable identity: \(a^2 - b^2\)
   • Trinomial of the form \(a^2 ± 2ab + b^2\) → Remarkable identity

Practice identifying patterns in algebraic expressions. The more you practice, the more intuitive recognition becomes!

Final tip: If you're unsure, start by looking for a common factor to extract before applying other methods.