How do you find the zeros from this equation

Mathematics

1 answer:

ozzi3 years ago

5 0

I recommend using the Rational root theorem and finding the zeros through synthetic division. The first zero is x = 1, so the first factorization is (x-1). That leaves a remainder of x^2 -3x-10. Now try to factor that. That factors out to (x-5)(x+2) so the factors of that cubic are (x-1)(x-5)(x+2)

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Roger and Tina took a test. Roger finished in 2 2/3 hours. Tina finished in 2 5/6 hours. Who finished first?

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4 years ago

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Question 9 (5 points)

Sunny_sXe [5.5K]

to get the equation of any straight line, we simply need two points off of it, let's use those two points from the table in the picture below.

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{13})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-17}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-17}-\stackrel{y1}{13}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-4)}}} \implies \cfrac{-30}{6 +4} \implies \cfrac{ -30 }{ 10 }\implies -3$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{13}=\stackrel{m}{-3}(x-\stackrel{x_1}{(-4)}) \\\\\\ y-13=-3(x+4)\implies y-13=-3x-12\implies y=-3x+1$