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Rewrite the expression as a product of four linear factors:

(12x^(2)-19 x)^(2)-5(12x^(2)-19 x)-50
Answer:

Rewrite the expression as a product of four linear factors: $\newline$ $\left(12 x^{2}-19 x\right)^{2}-5\left(12 x^{2}-19 x\right)-50$ $\newline$ Answer:

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Complex conjugate theorem

Full solution

Q. Rewrite the expression as a product of four linear factors:

\newline

\left(12 x^{2}-19 x\right)^{2}-5\left(12 x^{2}-19 x\right)-50

\newline

Answer:

Denote expression as $y$ : Let's denote the expression inside the parentheses as $y$ , so we have: $\newline$ $y = 12x^2 - 19x$ $\newline$ Now the given expression can be rewritten as: $\newline$ $y^2 - 5y - 50$ $\newline$ We will factor this quadratic expression in terms of $y$ .
Factor quadratic expression: To factor $y^2 - 5y - 50$ , we need to find two numbers that multiply to $-50$ and add up to $-5$ . These numbers are $-10$ and $5$ . So we can write the quadratic as: $(y - 10)(y + 5)$
Substitute back and simplify: Now we substitute back $12x^2 - 19x$ for $y$ to get the expression in terms of $x$ : $\newline$ $(12x^2 - 19x - 10)(12x^2 - 19x + 5)$
Factor first quadratic expression: Next, we need to factor each quadratic expression. We start with $12x^2 - 19x - 10$ . We look for two numbers that multiply to $12 \times -10 = -120$ and add up to $-19$ . These numbers are $-20$ and $6$ . So we can write the quadratic as: $(4x - 10)(3x + 1)$
Factor second quadratic expression: Now we factor the second quadratic expression, $12x^2 - 19x + 5$ . We look for two numbers that multiply to $12 \times 5 = 60$ and add up to $-19$ . These numbers are $-15$ and $-4$ . So we can write the quadratic as: $(3x - 5)(4x - 1)$
Write original expression as product: Finally, we write the original expression as a product of four linear factors: $(4x - 10)(3x + 1)(3x - 5)(4x - 1)$

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