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Factor completely: (6x-7)^(2)-(x+4)^(2) Answer:

Factor completely:\newline(6x−7)2−(x+4)2 (6 x-7)^{2}-(x+4)^{2} \newlineAnswer:

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Full solution

Q. Factor completely:\newline(6x−7)2−(x+4)2 (6 x-7)^{2}-(x+4)^{2} \newlineAnswer:
  1. Recognize as difference of squares: Recognize the expression as a difference of squares. A difference of squares is a mathematical expression of the form a2−b2a^2 - b^2, which can be factored into (a−b)(a+b)(a - b)(a + b).
  2. Identify 'a' and 'b': Identify 'a' and 'b' in the expression (6x−7)2−(x+4)2(6x-7)^2 - (x+4)^2. Here, 'a' is (6x−7)(6x-7) and 'b' is (x+4)(x+4).
  3. Apply formula to factor: Apply the difference of squares formula to factor the expression.\newlineUsing the formula (a2−b2)=(a−b)(a+b)(a^2 - b^2) = (a - b)(a + b), we get:\newline(6x−7)2−(x+4)2=(6x−7−(x+4))(6x−7+(x+4))(6x-7)^2 - (x+4)^2 = (6x-7 - (x+4))(6x-7 + (x+4))
  4. Simplify factored expression: Simplify the factored expression.\newlineNow we simplify each part:\newline(6x−7−x−4)=(5x−11)(6x-7 - x - 4) = (5x - 11)\newline(6x−7+x+4)=(7x−3)(6x-7 + x + 4) = (7x - 3)\newlineSo, the factored expression is (5x−11)(7x−3)(5x - 11)(7x - 3).

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