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Rewrite the expression as a product of four linear factors:

(x^(2)-6x)^(2)-11(x^(2)-6x)-80
Answer:

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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Complex conjugate theorem

Full solution

Q. 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:

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. $\newline$ 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. $\newline$ 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 $5$ . $\newline$ 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. $\newline$ We get $(x^2 - 6x - 16)(x^2 - 6x + 5)$ .
Factor Quadratics Separately: Next, we need to factor each quadratic separately. $\newline$ 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 $2$ . $\newline$ 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 $-1$ . $\newline$ 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. $\newline$ The expression $(x^2 - 6x)^2 - 11(x^2 - 6x) - 80$ can be rewritten as $(x - 8)(x + 2)(x - 5)(x - 1)$ .

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