Question

Show that (x – 3) is a factor of x³ – 7x² + 15x – 9. Hence factorise x³ – 7x² + 15 x – 9

Answer 1

Let x – 3 = 0, then x = 3,
Substituting the value of x in f(x),
f(x = x³ - 7x² + 15x – 9
= (3)³ – 7(3)² + 15(3) – 9
= 27 – 63 + 45 – 9
= 72 – 72
= 0
∵ Remainder = 0
∴ x – 3 is a factor of x³ – 7x² + 15x – 9
Now dividing it by x – 3, we get

`x - 3")"overline(x^3 - 7x^2 + 15x – 9)("x^2 - 4x + 3`
           x³ –  3x²              
            –   +                    
                  – 4x²  – 15x
                  – 4x² + 12x
                 +        –           
                           3x – 9
                           3x  – 9
                            –   +       
                                x        _
∴ x³ – 7x² + 15x – 9
= (x – 3)(x² –  4x + 3)
= (x – 3)[x² –  x – 3x + 3]
= (x – 3)[x(x – 1) –  3(x –  1)]
= (x – 3)(x – 1)(x –  3)
= (x – 3)² (x – 1).