Factor completely. (x^(2)-5x+4)(x^(2)-9)=

Factor the quadratic expression: Factor the quadratic expression (x^2 - 5x + 4). We look for two numbers that multiply to 4 and add up to -5. These numbers are -4 and -1. So, we can factor (x^2 - 5x + 4) as (x - 4)(x - 1).Factor the difference of squares: Factor the difference of squares (x^2 - 9). We recognize that 9 is a perfect square, and the expression is in the form a^2 - b^2, which factors into (a + b)(a - b). Here, a is x and b is 3, so (x^2 - 9) factors into (x + 3)(x - 3).Combine the factors: Combine the factors from Step 1 and Step 2 to write the complete factorization. The complete factorization of the given expression is (x - 4)(x - 1)(x + 3)(x - 3).

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Algebra 2

Factor sums and differences of cubes

Full solution

Q. Factor completely.

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\left(x^{2}-5 x+4\right)\left(x^{2}-9\right)=

Factor the quadratic expression: Factor the quadratic expression $(x^2 - 5x + 4)$ . $\newline$ We look for two numbers that multiply to $4$ and add up to $-5$ . These numbers are $-4$ and $-1$ . $\newline$ So, we can factor $(x^2 - 5x + 4)$ as $(x - 4)(x - 1)$ .
Factor the difference of squares: Factor the difference of squares $(x^2 - 9)$ . $\newline$ We recognize that $9$ is a perfect square, and the expression is in the form $a^2 - b^2$ , which factors into $(a + b)(a - b)$ . $\newline$ Here, $a$ is $x$ and $b$ is $3$ , so $(x^2 - 9)$ factors into $(x + 3)(x - 3)$ .
Combine the factors: Combine the factors from Step $1$ and Step $2$ to write the complete factorization. $\newline$ The complete factorization of the given expression is $(x - 4)(x - 1)(x + 3)(x - 3)$ .