Using factor theorem, show that (x - 4) is a factor of (2x³ + x² - 26x - 40) and hence factorize (2x³ + x² - 26x - 40).

Let f(x) = 2x³ + x² - 26x - 40

Substituting x = 4 in f(x), we get :

f(4) = 2(4)³ + (4)² - 26(4) - 40

= 2(64) + 16 - 104 - 40

= 128 + 16 - 104 - 40

= 144 - 144

= 0.

Since f(4) = 0, (x − 4) is a factor of 2x³ + x² - 26x - 40

Now, dividing f(x) by x - 4,

$\begin{array}{l} \phantom{x - 3)}{2x^2 + 9x + 10} \ x - 4\overline{\smash{\big)}2x^3 + x^2 - 26x - 40} \ \phantom{x - 2}\underline{\underset{-}{+}2x^3 \underset{+}{-}8x^2} \ \phantom{{x - 2}x^3-}9x^2 - 26x \ \phantom{{x -l2}x^3\space}\underline{\underset{-}{+}9x^2\underset{+}{-} 36x} \ \phantom{{x - 2uo[ki]}x^3okk\space}{10x - 40} \ \phantom{{x - 2}x^3o;lmkk\space}\underline{\underset{-}{+}10x\underset{+}{-} 40} \ \phantom{{x - 2}{x^,j-k2x^2\space}{-9x}}\times \end{array}$

∴ 2x³ + x² - 26x - 40 = (x - 4)(2x² + 9x + 10)

= (x - 4)(2x² + 4x + 5x + 10)

= (x - 4)[2x(x + 2) + 5(x + 2)]

= (x - 4)(2x + 5)(x + 2)

Hence, 2x³ + x² - 26x - 40 = (x - 4)(2x + 5)(x + 2).