Using factor theorem, show that (x - 3) is a factor of (x³ - 7x² + 15x - 9). Hence, factorize the given expression completely.

Let f(x) = x³ - 7x² + 15x - 9.

We know that,

(x - 3) will be the factor of f(x), if f(3) will be equal to 0.

⇒ f(3) = (3)³ - 7(3)² + 15(3) - 9.

= 27 - 63 + 45 - 9

= 72 - 72

= 0.

Since, f(3) = 0, thus (x - 3) is a factor of f(x).

Now, dividing f(x) by x - 3,

$\begin{array}{l} \phantom{x - 3)}{x^2 - 4x + 3} \ x - 3\overline{\smash{\big)}x^3 - 7x^2 + 15x - 9} \ \phantom{x - 2}\underline{\underset{-}{}x^3 \underset{+}{-}3x^2} \ \phantom{{x - 2}x^3-}-4x^2 + 15x \ \phantom{{x - 2}x^3ok\space}\underline{\underset{+}{-}4x^2\underset{-}{+} 12x} \ \phantom{{x - 2uo[ki]}x^3okklk\space}{3x - 9} \ \phantom{{x - 2}x^3o;lmkb\k\space}\underline{\underset{-}{+}3x\underset{+}{-} 9} \ \phantom{{x - 2}{x^,jo\3-2x^2\space}{-9x}}\times \end{array}$

∴ x³ - 7x² + 15x - 9 = (x - 3)(x² - 4x + 3)

= (x - 3)(x² - 3x - x + 3)

= (x - 3)[x(x - 3) - 1(x - 3)]

= (x - 3)(x - 1)(x - 3)

= (x - 3)²(x - 1).

Hence, x³ - 7x² + 15x - 9 = (x - 3)²(x - 1).