91Ó°ÊÓ

1. First of all, guess a simple root for p(x). 2. Use synthetic division to divide p(x) by (x-root). 3. If the remainder is 0, it means p(x) has a factor (x-root). 4. Repeat the process with the remaining factors until we reach a quadratic polynomial. 5. Once you have a quadratic polynomial, determine the remaining roots. Here's a step-by-step process to factor p(x): 1. By trying out some integer values of x, we find that 1 is a root of p(x) because p(1) = 0. 2. Now, use synthetic division to divide p(x) by (x-1): ```plaintext x^2 x 1 ________________________________ x-1 | x^3 0 -2x -1 x^3 -x^2 ______________________________ x^2 -2x x^2 -x _______________________________ 0 -1 ``` 3. The remainder is 0, so p(x) has a factor of (x-1). The quotient is x^2 + x + 1. 4. We now have p(x) = (x - 1)(x^2 + x + 1). 5. The remaining quadratic factor (x^2 + x + 1) has no rational roots, so p(x) is completely factored in this form. Hence, the factored form of p(x) is \((x - 1)(x^2 + x + 1)\).

1. We try to find a simple root for q(x). 2. After trying out some integer values of x, we find that 1 and 2 are roots because q(1) = 0 and q(2) = 0. 3. Use synthetic division to divide q(x) by (x-1) and then (x-2). If the remainder is 0, it means q(x) has those factors. 4. Keep repeating the process with the remaining factors until we reach a quadratic polynomial. 5. Once you have a quadratic polynomial, determine the remaining roots. Step-by-Step process to factor q(x): 1. We know that 1 and 2 are roots of q(x). 2. First, use synthetic division to divide q(x) by (x-1): ```plaintext x^3 -2x^2 x 6 ________________________________ x-1 | x^4 -3x^3 -3x^2 +11x -6 x^4 -x^3 ______________________________ -2x^3 -3x^2 -2x^3 +2x^2 _______________________________ -x^2 +11x -x^2 +x _______________________________ 10x -6 10x -10 _______________________________ 4 ``` Since the remainder is not 0, we can say that q(x) does not have a factor (x-1). Now let's try synthetic division with (x-2): ```plaintext x^3 x^2 5x -3 ________________________________ x-2 | x^4 -3x^3 -3x^2 +11x -6 x^4 -2x^3 ______________________________ -1x^3 -3x^2 -1x^3 2x^2 _______________________________ -1x^2 +11x -1x^2 +2x _______________________________ 9x -6 9x -18 _______________________________ 12 ``` Since the remainder is not 0, we can say that q(x) does not have a factor (x-2) either. At this point, we have exhausted all integer possibilities for the roots of this polynomial. We must resort to other methods to find the roots, such as factoring by grouping or by using a graphing calculator. However, this exercise requires the solutions in linear factor form, and since we were unable to find roots by inspection, we cannot provide an answer in that format.

To factor r(x), first notice that it looks like a quadratic equation in x^2. We can let y = x^2 to make it look like y^2 - 6y + 7 1. Factor the quadratic equation: y^2 - 6y + 7 = (y-1)(y-7) 2. Now, replace y with x^2: r(x) = (x^2 - 1)(x^2 - 7) 3. Further factor the first term (x^2 - 1) into linear factors: (x + 1)(x - 1) 4. Finally, express r(x) as a product of linear factors: r(x) = (x + 1)(x - 1)(x^2 - 7) Unfortunately, r(x) cannot be broken down further into linear factors with rational coefficients, so this is our final answer. Hence, the factored form of r(x) is \((x + 1)(x - 1)(x^2 - 7)\).