Rewrite the expression as a product of four linear factors: (x^(2)-6x)^(2)-11(x^(2)-6x)-80 Answer:

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Rewrite the expression as a product of four linear factors:  (x^(2)-6x)^(2)-11(x^(2)-6x)-80 Answer:

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Denote and Substitute: Let's denote $ u = x^2 - 6x $. We can then rewrite the given expression in terms of $ u $ to simplify the problem. Substitute $ u $ into the expression: $ (u)^2 - 11u - 80 $.Factor Quadratic: Now we have a quadratic in terms of $ u $: $ u^2 - 11u - 80 $. We need to factor this quadratic. To factor $ u^2 - 11u - 80 $, we look for two numbers that multiply to -80 and add up to -11. These numbers are -16 and 5. So, $ u^2 - 11u - 80 $ factors into $ (u - 16)(u + 5) $.Substitute Back and Factor: Now we substitute back $ x^2 - 6x $ for $ u $ in the factored form. We get $ (x^2 - 6x - 16)(x^2 - 6x + 5) $.Factor Quadratics Separately: Next, we need to factor each quadratic separately. Starting with $ x^2 - 6x - 16 $, we look for two numbers that multiply to -16 and add up to -6. These numbers are -8 and 2. So, $ x^2 - 6x - 16 $ factors into $ (x - 8)(x + 2) $.Factor Second Quadratic: Now, we factor $ x^2 - 6x + 5 $. We look for two numbers that multiply to 5 and add up to -6. These numbers are -5 and -1. So, $ x^2 - 6x + 5 $ factors into $ (x - 5)(x - 1) $.Write Original Expression: Finally, we write the original expression as a product of four linear factors using the factors we found. The expression $ (x^2 - 6x)^2 - 11(x^2 - 6x) - 80 $ can be rewritten as $ (x - 8)(x + 2)(x - 5)(x - 1) $.

Rewrite the expression as a product of four linear factors: $\newline$ $\left(x^{2}-6 x\right)^{2}-11\left(x^{2}-6 x\right)-80$ $\newline$ Answer:

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Rewrite the expression as a product of four linear factors:\newline(x2−6x)2−11(x2−6x)−80 \left(x^{2}-6 x\right)^{2}-11\left(x^{2}-6 x\right)-80 (x2−6x)2−11(x2−6x)−80\newlineAnswer:

Rewrite the expression as a product of four linear factors: $\newline$ $\left(x^{2}-6 x\right)^{2}-11\left(x^{2}-6 x\right)-80$ $\newline$ Answer: