Question

(y-1) हा (my-2y² +y³-4) बहूपदिचा अवयव असेल तर m ची किमत काढा
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Answer 1

Answer....

$\tt{y - 1} = 0 \\ \tt y = 1 \\ \\ \tt = (my \: - {2y}^{2} + {y}^{2} - 4) = 0\\ \tt = m \times 1 - 2 {(1)}^{2} + {(1)}^{2} - 4 = 0 \\ \tt = m - 2 + 1 = 0 \\ \tt = m - 1 = 0 \\ \tt = m = 1$