Question

show that 2 , -1 and 1/2 are zeros of the cubic polynomial p(x)2x^3-3x^2 -3x+2​

Answer 1

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$put \: x = 2$

$2(2) {}^{3} - 3(2) {}^{2} - 3(2) + 2 = 0 \\ \\ 16 - 12 - 6 + 2 = 0 \\ \\ 16 - 12 - 4 = 0\\ \\ 4 - 4 = 0 \\ 0 = 0$

$put \: x = - 1$

$2( - 1) {}^{3} - 3( - 1) {}^{2} - 3( - 1) + 2 = 0 \\ \\ - 2 - 3 + 3 + 2 = 0 \\ \\ - 5 + 5 = 0 \\ \\ 0 = 0$

$put \: x = \frac{1}{2} \\ \\ 2 (\frac{1}{2} ) {}^{3} - 3(\frac{1}{2} ) {}^{2} - 3( \frac{1}{2} ) + 2 = 0\\ \\ \frac{2}{8} - \frac{3}{4} - \frac{3}{2} + 2 = 0 \\ \\ \\ \frac{2 - 6 - 12 + 16}{8} = 0\\ \\ \frac{18 - 18}{8} = 0 \\ \\ \frac{0}{8} = 0 \\ \\ 0 = 0$

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Answer 2

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