Question

4r²+8r.find 2 zeros​

Answer 1

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$\rightarrow 4 {r}^{2} + 8r. \\ \\ \rightarrow \: 4r(r + 2). \\ \\ \rightarrow \: 4r = 0. \: \: \: \: \: \: \: \: \: \: and \: \: \: \: \: \: \: \: r + 2 = 0 \\ \\ \rightarrow \: r = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: and \: \: \: \: \: r = - 2.$

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Answer 2

4r^2 + 8r = 0

4r (r + 2) =0

4r =0 , r + 2 = 0

r =0, -2

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