how do you factor completely 3(x - 2)^2 - 3(x - 2) - 6
That can be recognized as a polynomial in x-2. It factors into
[3(x-2)+3][(x-2) -2]
which simplifies to
= (3x+3)(x-4) = 3(x+1)(x-4)
Another way to do that would be to multiply out the polynomial 3(x-2)^2 -3(x-2) -6 BEFORE factoring it. The answer should be the same.
To factor completely the expression 3(x - 2)^2 - 3(x - 2) - 6, you can follow these steps:
Step 1: Identify any common factors among the terms in the expression, if possible. In this case, we see that all the terms have a common factor of 3. We can factor out 3 from each term:
3[(x - 2)^2 - (x - 2) - 2]
Step 2: Focus on the expression inside the brackets: (x - 2)^2 - (x - 2) - 2. We can simplify this further by expanding the square.
(x - 2)^2 can be expanded to (x - 2)(x - 2), which gives us:
(x - 2)(x - 2) - (x - 2) - 2
Step 3: Simplify the expression inside the brackets.
[(x - 2)(x - 2)] - (x - 2) - 2
(x - 2)(x - 2 - 1) - 2
(x - 2)(x - 3) - 2
Step 4: Bringing back the common factor of 3, we rewrite the expression as:
3(x - 2)(x - 3) - 2
Therefore, the completely factored form of the expression 3(x - 2)^2 - 3(x - 2) - 6 is 3(x - 2)(x - 3) - 2.