Problem 85 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$x^{2}-12 x+36-49 y^{2}$$

Step-by-Step Solution

Answer

The completely factored form of the polynomial \(x^{2}-12 x+36-49 y^{2}\) is \((x-6-7y)(x-6+7y)\)

Step 1: Identifying the structure of the polynomial

The first thing to notice is the structure of the polynomial. The polynomial \(x^{2}-12 x+36-49 y^{2}\) can be written as \((x^2 -12x + 36) - (7y)^2\). In other words, there are two terms, the first is a perfect square trinomial and the second is the square of 7y.

Step 2: Factoring the perfect square trinomial

Next, factor the perfect square trinomial \(x^2 - 12x + 36\). This trinomial is equivalent to \((x - 6)^2\). So now the polynomial is re-written as \((x-6)^2 - (7y)^2\).

Step 3: Applying the difference of squares formula

Then apply the difference of squares formula, which states that \(a^2-b^2 = (a-b)(a+b)\). Substituting \(a=(x-6)\) and \(b=7y\) into this formula gives \((x-6-7y)(x-6+7y)\) as the completely factored form of the original polynomial.