Rewrite the expression as a product of four linear factors: (x^(2)-3x)^(2)-22(x^(2)-3x)+72 Answer:

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Rewrite the expression as a product of four linear factors:  (x^(2)-3x)^(2)-22(x^(2)-3x)+72 Answer:

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Identify and Set Up: Let's identify the given expression and set it up for factoring: The expression is a quadratic in terms of (x^2 - 3x), which can be represented as: Let y = x^2 - 3x Then the expression becomes: y^2 - 22y + 72 Now we will factor this quadratic equation.Find Factors: We need to find two numbers that multiply to 72 and add up to -22. These numbers are -18 and -4 because -18 * -4 = 72 and -18 + -4 = -22. So we can factor the quadratic as: (y - 18)(y - 4)Substitute and Simplify: Now we substitute back x^2 - 3x for y to get: (x^2 - 3x - 18)(x^2 - 3x - 4)Factor x^2 - 3x - 18: Next, we need to factor each quadratic expression further. Starting with x^2 - 3x - 18, we look for two numbers that multiply to -18 and add up to -3. These numbers are -6 and +3. So we can factor x^2 - 3x - 18 as: (x - 6)(x + 3)Factor x^2 - 3x - 4: Now, we factor x^2 - 3x - 4 by finding two numbers that multiply to -4 and add up to -3. These numbers are -4 and +1. So we can factor x^2 - 3x - 4 as: (x - 4)(x + 1)Write as Product: Finally, we write the original expression as a product of four linear factors: (x - 6)(x + 3)(x - 4)(x + 1)

Rewrite the expression as a product of four linear factors: $\newline$ $\left(x^{2}-3 x\right)^{2}-22\left(x^{2}-3 x\right)+72$ $\newline$ Answer:

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Rewrite the expression as a product of four linear factors:\newline(x2−3x)2−22(x2−3x)+72 \left(x^{2}-3 x\right)^{2}-22\left(x^{2}-3 x\right)+72 (x2−3x)2−22(x2−3x)+72\newlineAnswer:

Rewrite the expression as a product of four linear factors: $\newline$ $\left(x^{2}-3 x\right)^{2}-22\left(x^{2}-3 x\right)+72$ $\newline$ Answer: