Problem 74 - Algebra and Trigonometry

Question

Factor completely, or state that the polynomial is prime. $$6 x^{2}-6 x-12$$

Step-by-Step Solution

Answer

The completely factorized form of the polynomial \(6x^{2}-6x-12\) is \(6(x-2)(x+1)\)

Step 1: Identify Common Factor

Every term in the polynomial \(6x^{2}-6x-12\) has a common factor of 6. The first step is to factor out this common factor.

Step 2: Factor out the Common Factor

By factoring out the common factor of 6, it gives: \(6(x^{2}-x-2)\)

Step 3: Factorize the Quadratic Polynomial

Now, the quadratic polynomial \((x^{2}-x-2)\) needs to be factorized. By looking at the terms, one can identify that it can be factored into \((x-2)(x+1)\) as \(-2x + x -2\), which is equal to \(-x -2\). Thus, the quadratic polynomial can be factorized as: \(x^{2}-x-2 = (x-2)(x+1)\)

Step 4: Write Down the Final Factorized Form

Now replace the factorized quadratic polynomial from step 3 into the expression from step 2 which gives: \(6(x^{2}-x-2) = 6(x-2)(x+1)\)