Question

find the zeros of the cubic polynomial x3minus 7x2 plus 36,g

Answer 1

${x}^{3} - 7 {x}^{2} + 36 = 0 \\ {x}^{3} - 6 {x}^{2} - {x}^{2} + 36 = 0 \\ {x}^{2} (x - 6) - 1(x - 6)(x + 6) = 0 \\ (x - 6)( {x}^{2} - x - 6) = 0 \\ (x - 6)( {x}^{2} - 3x + 2x - 6) = 0 \\ (x - 6)(x(x - 3) + 2(x - 3)) = 0 \\ (x - 6)(x - 3)(x + 2) = 0 \\ therefore \: zeroes \: are \\ x = 6 \: \: and \: 3 \: \: and \: - 2$