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Factor completely: (7x+1)^(2)-(4x+3)^(2) Answer:

Factor completely:\newline(7x+1)2−(4x+3)2 (7 x+1)^{2}-(4 x+3)^{2} \newlineAnswer:

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Full solution

Q. Factor completely:\newline(7x+1)2−(4x+3)2 (7 x+1)^{2}-(4 x+3)^{2} \newlineAnswer:
  1. Recognize as difference of squares: Recognize the expression as a difference of squares.\newlineThe given expression is in the form of a2−b2a^2 - b^2, which is a difference of squares.\newlinea=(7x+1)a = (7x+1) and b=(4x+3)b = (4x+3).\newlineDifference of squares formula: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b).
  2. Apply formula: Apply the difference of squares formula.\newlineUsing the formula from Step 11, we can write the expression as:\newline(7x+1+4x+3)(7x+1−4x−3)(7x+1 + 4x+3)(7x+1 - 4x-3).
  3. Simplify terms: Simplify the terms inside the parentheses.\newlineSimplify the first set of parentheses by adding the like terms:\newline(7x+4x)+(1+3)=11x+4(7x + 4x) + (1 + 3) = 11x + 4.\newlineSimplify the second set of parentheses by subtracting the like terms:\newline(7x−4x)+(1−3)=3x−2(7x - 4x) + (1 - 3) = 3x - 2.
  4. Write final form: Write the final factored form.\newlineThe completely factored form of the expression is:\newline(11x+4)(3x−2)(11x + 4)(3x - 2).

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