Question

x^3-x^2-14x+24------factorise

Answer 1

     P(x) = x³ - x² - 14 x + 24

     factors of 24 are possible rational roots of P(x), if at all rational roots exist.

     so roots are possibly : + 1, 2, 3, 4, 6,8, 12, 24  or -1, -2, -3, -4, -6, -8, -12, -24

   check with 1, -1, 2 ..  x = 2 => P(2) = 0

         let,   P(x) = (x - 2) [ x² - a x + (24/-2)  ]

             then compare coefficients of x² term or x term, then
        
           -a x² - 2 x²  = - 1 x²      =>  a = -1
           2 a x -12 x = - 14 x    =>  a = -1

       P(x) =  ( x -2 ) ( x² +x-12) =
        
           find factors of -12 that have difference of  -1.
               so  -4 and 3 are roots of x² +x -12

       P(x) = (x - 2 ) ( x + 4 ) ( x - 3)

Answer 2

Answer:

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