Question

verify that 2,3 and -1 are the zeroes of the cubic polynomial x^3-4x^2+x+6​

Answer 1

Answer:

${x}^{3} - 4 {x}^{2} + x + 6 = 0 \\ f(2) = {2}^{3} - 4 ({2})^{2} + 2 + 6 \\ = 8 - 16 + 2 + 6 \\ = f(3) = {3}^{3 } - 4( {3}^{2} ) + 3 + 6 \\ = 27 + - 36 + 3 + 6 \\ = 0 \\ f( - 1) = { - 1}^{3} - 4( { - 1}^{2} ) - 1 + 6 \\ = - 1 - 4 - 1 + 6 = 0 \\ therefore \: 2 \: 3 \: ( - 1) \: are \: the \: roots \: of \: the \: equatio$

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