Using factor theorem, factorize the following:

x³ + 7x² + 7x - 15

Let, f(x) = x³ + 7x² + 7x - 15.

Substituting, x = 1 in f(x), we get :

f(1) = (1)³ + 7(1)² + 7(1) - 15

= 1 + 7 + 7 - 15

= 0.

Since, f(1) = 0, thus (x - 1) is a factor of f(x).

Dividing, f(x) by (x - 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{x^2 + 8x + 15} \ x - 1\overline{\smash{\big)}x^3 + 7x^2 + 7x - 15} \ \phantom{x - 2l}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \ \phantom{{x - 2}x^,,,3-}8x^2 + 7x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}8x^2 \underset{+}{-} 8x} \ \phantom{{x - 2]euo[ki]}x^3okk\space}{15x - 15} \ \phantom{{x - 2}x^3o;llk]lmk\space}\underline{\underset{-}{+}15x\underset{+}{-} 15} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ x³ + 7x² + 7x - 15 = (x - 1)(x² + 8x + 15)

= (x - 1)(x² + 3x + 5x + 15)

= (x - 1)[x(x + 3) + 5(x + 3)]

= (x - 1)(x + 5)(x + 3).

Hence, x³ + 7x² + 7x - 15 = (x - 1)(x + 5)(x + 3).