Use Remainder Theorem to factorise the following polynomials completely :

(i) 2x³ + x² - 13x + 6

(ii) 3x³ + 2x² - 19x + 6

(iii) 2x³ + 3x² - 9x - 10

(iv) x³ + 10x² - 37x + 26

(i) Let f(x) = 2x³ + x² - 13x + 6

Putting, x = 2 in f(x)

$f(2) = 2(2)^3 + 2^2 - 13(2) + 6 \\[0.5em] = 16 + 4 - 26 + 6 \\[0.5em] = 0$

Since, f(2) = 0 , (x - 2) is factor of f(x) by factor theorem.
Dividing, f(x) by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 5x - 3} \ x - 2\overline{\smash{\big)}2x^3 + x^2 - 13x + 6} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{+}{-} 4x^2} \ \phantom{{x - 2}2x^3+4}5x^2 - 13x \ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}5x^2 \underset{+}{-} 10x} \ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, 2x² + 5x - 3 as quotient and remainder = 0.

$\therefore 2x^3 + x^2 - 13x + 6 = (x - 2)(2x^2 + 5x - 3) \\[0.5em] = (x - 2)(2x^2 + 6x - x - 3) \\[0.5em] = (x - 2)(2x(x + 3) - 1(x + 3)) \\[0.5em] = (x - 2)(2x - 1)(x + 3)$

Hence, 2x³ + x² - 13x + 6 = (x - 2)(2x - 1)(x + 3).

(ii) Let f(x) = 3x³ + 2x² - 19x + 6

Putting, x = 2 in f(x)

$f(2) = 3(2)^3 + 2(2)^2 - 19(2) + 6 \\[0.5em] = 24 + 8 - 38 + 6 \\[0.5em] = 38 - 38 \\[0.5em] = 0$

Since, f(2) = 0, (x - 2) is factor of f(x) by factor theorem.

Dividing, f(x) by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{3x^2 + 8x - 3} \ x - 2\overline{\smash{\big)}3x^3 + 2x^2 - 19x + 6} \ \phantom{x - 2}\underline{\underset{-}{}3x^3 \underset{+}{-} 6x^2} \ \phantom{{x - 2}2x^3+4}8x^2 - 19x \ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}8x^2 \underset{-}{+} 16x} \ \phantom{{x - 2}{2x^3+}{+2x^2}}-3x + 6 \ \phantom{{x - 2}{2x^3+}{+2x^2}{4}}\underline{\underset{+}{-}3x \underset{-}{+} 6} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, 3x² + 8x - 3 as quotient and remainder = 0.

$\therefore 3x^3 + 2x^2 - 19x + 6 = (x - 2)(3x^2 + 8x - 3) \\[0.5em] = (x - 2)(3x^2 + 9x - x - 3) \\[0.5em] = (x - 2)(3x(x + 3) - 1(x + 3)) \\[0.5em] = (x - 2)(3x - 1)(x + 3)$

Hence, 3x³ + 2x² - 19x + 6 = (x - 2)(3x - 1)(x + 3).

(iii) Let f(x) = 2x³ + 3x² - 9x - 10

Putting, x = 2 in f(x)

$f(2) = 2(2)^3 + 3(2)^2 - 9(2) - 10 \\[0.5em] = 16 + 12 - 18 - 10 \\[0.5em] = 28 - 28 \\[0.5em] = 0$

Since, f(2) = 0, (x - 2) is factor of f(x) by factor theorem.

Dividing, f(x) by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 7x + 5} \ x - 2\overline{\smash{\big)}2x^3 + 3x^2 - 9x - 10} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{-}{+}4x^2} \ \phantom{{x - 2}2x^3+4}7x^2 - 9x \ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}7x^2 \underset{+}{-} 14x} \ \phantom{{x - 2}{2x^3+}{+2x^2}}5x - 10 \ \phantom{{x - 2}{2x^3+}{+2x}}\underline{\underset{-}{ }5x \underset{+}{-} 10} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, 2x² + 7x + 5 as quotient and remainder = 0.

$\therefore 2x^3 + 3x^2 - 9x - 10 = (x - 2)(2x^2 + 7x + 5) \\[0.5em] = (x - 2)(2x^2 + 5x + 2x + 5) \\[0.5em] = (x - 2)(x(2x + 5) + 1(2x + 5)) \\[0.5em] = (x - 2)(x + 1)(2x + 5)$

Hence, 2x³ + 3x² - 9x - 10 = (x - 2)(x + 1)(2x + 5).

(iv) Let f(x) = x³ + 10x² - 37x + 26

Putting, x = 1 in f(x)

$f(1) = (1)^3 + 10(1)^2 - 37(1) + 26 \\[0.5em] = 1 + 10 - 37 + 26 \\[0.5em] = 37 - 37 \\[0.5em] = 0$

Since, f(1) = 0 , (x - 1) is factor of f(x) by factor theorem.

Dividing, f(x) by (x - 1),

$\begin{array}{l} \phantom{x - 1)}{x^2 + 11x - 26} \ x - 1\overline{\smash{\big)}x^3 + 10x^2 - 37x + 26} \ \phantom{x - 1}\underline{\underset{-}{}x^3 \underset{+}{-}x^2} \ \phantom{{x - 1}2x^3+4}11x^2 - 37x \ \phantom{{x - 1}2x^3+}\underline{\underset{-}{}11x^2 \underset{+}{-} 11x} \ \phantom{{x - 1}{2x^3+}{+11x^2}}-26x + 26 \ \phantom{{x - 1}{2x^3+}{+11x^2}}\underline{\underset{+}{-}26x \underset{-}{+} 26} \ \phantom{{x - 1}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, x² + 11x - 26 as quotient and remainder = 0.

$\therefore x^3 + 10x^2 - 37x + 26 = (x - 1)(x^2 + 11x - 26) \\[0.5em] = (x - 1)(x^2 + 13x - 2x - 26) \\[0.5em] = (x - 1)(x(x + 13) - 2(x + 13)) \\[0.5em] = (x - 1)(x - 2)(x + 13)$

Hence, x³ + 10x² - 37x + 26 = (x - 1)(x - 2)(x + 13).