Using remainder theorem factorise 4x³ + 7x² - 36x - 63 completely.

Substituting x = 3 in 4x³ + 7x² - 36x - 63, we get :

⇒ 4x³ + 7x² - 36x - 63

⇒ 4(3)³ + 7(3)² - 36(3) - 63

⇒ 4 × 27 + 7 × 9 - 108 - 63

⇒ 108 + 63 - 108 - 63

⇒ 0.

So, x - 3 is the factor of 4x³ + 7x² - 36x - 63.

On dividing 4x³ + 7x² - 36x - 63 by (x - 3) we get,

$\begin{array}{l} \phantom{x - 3}{4x^2 + 19x + 21} \ x - 3 \overline{\smash{\big)}4x^3 + 7x^2 - 36x - 63} \ \phantom{x - 3}\underline{\underset{-}{}4x^3 \underset{+}{-}12x^2} \ \phantom{{x - 3}4x^3+1}19x^2 - 36x \ \phantom{{x - 3}x^3+1}\underline{\underset{-}{}19x^2 \underset{+}{-}57x} \ \phantom{{x - 3}4x^3-12x^2+5}21x - 63 \ \phantom{{x - 3}4x^3-12x^2+}\underline{\underset{-}{}21x\underset{+}{-}63} \ \phantom{{x - 3}4x^3-12x^2+5333} \times \end{array}$

∴ 4x³ + 7x² - 36x - 63 = (x - 3)(4x² + 19x + 21)

= (x - 3)(4x² + 12x + 7x + 21)

= (x - 3)[4x(x + 3) + 7(x + 3)]

= (x - 3)(4x + 7)(x + 3).

Hence, 4x³ + 7x² - 36x - 63 = (x - 3)(4x + 7)(x + 3).