x^3-4x^2-7x+28=0

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^{x^3-4x^2-7x+28=0}

^{What are the numbers of possible number of positive real solutions?}

^{A. Two or zero}

^{B. Three or one}

^{C. Two}

^{D. One}

Guest Jan 27, 2020

geno3141 · Answer 1 · 2020-01-27T09:54:42

x³ - 4x² - 7x + 28 = 0

Factor x² out of the first two terms:

x²(x - 4) - 7x + 28 = 0

Factor -7 out of the last two terms:

x²(x - 4) - 7(x - 4) = 0

Factor x - 4 out of these two terms:

(x - 4)(x² - 7) = 0

Factor x² - 7:

(x - 4)(x - sqrt(7))(x + sqrt(7)) = 0

This shows that there are 3 real answers.