Question

show that (x-2),(x+3) and (x-7) are factors of x³- 6x²-13x +42

Answer 1

if (x-2) is a factor
f(2) =0 (just put 2 in place of x)
f(-3) = 0
f(7) =0
if all solution give zero then (x-2), (x+3),(x-7) are the factors

Answer 2

${x}^{3} - 6 {x}^{2} - 13x + 42 \\ {x}^{3} - 2 {x}^{2} - 4 {x}^{2} + 8x - 21x + 42 \\ {x}^{2} (x - 2) - 4x(x - 2) - 21(x - 2) \\ (x - 2)( {x}^{2} - 4x - 21) \\ (x - 2)( {x}^{2} - 7x + 3x - 21) \\ (x - 2)(x - 7)(x + 3)$
i did vanishing factor taking (x-2) then found the
$(x - 7)( {x}^{2} - 4x - 21)$
the by middle term factor i proved the given question