Question

Find the number of real zeroes in p(x) = (x+1)(x-2)(x^2 +3)

Answer 1

solution:

(x+1)(x-2)(x^2+3)=0

=> x+1=0

=> ✔x= 1✔

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=> x -2=0

=> ✔X= 2 ✔

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${x}^{2} + 3 = 0 \\ = > x = \sqrt{ - 3}$

But, this is not real value of x.

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so, x has only two real value.

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Hope it helps:)

#Amrit⭐

Answer 2

☆hey friend!!!☆

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here is your answer ☞
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$given \: \\ p(x) \: = \: (x + 1)(x - \: \: 2)( {x}^{2} + 3) \\ \\ now \\ \\ = > \: p(x) \: = \: 0 \\ \\ = > \: (x + 1)(x - 2)( {x}^{2} + 3) \: = \: 0 \\ \\ = > x = - 1. \\ = > x = \: 2 \\ = > x \: = \sqrt{-3}$
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hope it will help you ☺☺☺☺
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Devil_king ▄︻̷̿┻̿═━一