Using Factor Theorem, show that :

(x + 5) is a factor of 2x³ + 5x² - 28x - 15. Hence, factorise the expression 2x³ + 5x² - 28x - 15 completely.

x + 5 = 0 ⇒ x = -5.

Remainder = The value of 2x³ + 5x² - 28x - 15 at x = -5.

= 2(-5)³ + 5(-5)² - 28(-5) - 15

= 2(-125) + 5(25) + 140 - 15

= -250 + 125 + 140 - 15

= -265 + 265

= 0.

Hence, (x + 5) is a factor of 2x³ + 5x² - 28x - 15.

Now dividing 2x³ + 5x² - 28x - 15 by (x + 5),

$\begin{array}{l} \phantom{x + 5)}{2x^2 - 5x - 3} \ x + 5\overline{\smash{\big)}2x^3 + 5x^2 - 28x - 15} \ \phantom{x - 5}\underline{\underset{-}{}2x^3 \underset{-}{+}10x^2} \ \phantom{{x - 5}2x^3+}-5x^2 - 28x \ \phantom{{x - 5}2x^3+}\underline{\underset{+}{-}5x^2 \underset{+}{-} 25x} \ \phantom{{x - 5}{2x^3+}{+5x^2+}}-3x - 15 \ \phantom{{x - 5}{2x^3+}{+5x^2+\enspace}}\underline{\underset{+}{-}3x \underset{+}{-} 15} \ \phantom{{x - 5}{2x^3+}{+5x^2+\enspace}{-3x}}\times \end{array}$

we get quotient = 2x² - 5x - 3

Factorising 2x² - 5x - 3,

⇒ 2x² - 6x + x - 3

⇒ 2x(x - 3) + 1(x - 3)

⇒ (2x + 1)(x - 3)

∴ 2x³ + 5x² - 28x - 15 = (x + 5)(2x + 1)(x - 3).

Hence, 2x³ + 5x² - 28x - 15 = (x + 5)(2x + 1)(x - 3).