Question

Factorise:
${x}^{3}-23 {x}^{2} + 142x - 120$
​

Answer 1

$\huge\underline\mathfrak{Question-}$

$= > x^{3} - 23x^{2} + 124x - 120$

$\huge\underline\mathfrak{Answer-}$

$= > (x - 12)(x - 10)(x - 1)$

$\small\underline\mathfrak{Step\:By\:Step\: Explanation\:}$-

$= > {x}^{3} - {23x}^{2} + 22x + 120x - 120$

$= > {x}^{3} - 23x^{2} + 22x + 120x - 120$

$= > {x}^{2} (x - 1) - 22x(x - 1) + 120(x - 1)$

$= > ( {x}^{2} - 22x + 120)(x - 1)$

$= > (x^{2} - 10x - 12x + 120)(x - 1)$

$= > (x(x - 10) - 12(x - 10))(x - 1)$

$= > (x - 12)(x - 10)(x - 1)$

Answer 2

Solutions :-


$Let \: \:p(x) \: {x}^{3}-23 {x}^{2} + 142x - 120$

Factors of - 120 = ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60

Taking the value of x = 1

$p(1) \: {1}^{3}-23 \times {1}^{2} + 142 \times 1 - 120 \\ \\ = > 1 - 23 + 142 - 120 = 0$
So,
x = 1 => x - 1 = 0


$Now, \\ Divide \: \: {x}^{3}-23 {x}^{2} + 142x - 120 \: \: by \: \: x - 1 \\ \\ We\: \: get, \\ \\ = > \frac{{x}^{3}-23 {x}^{2} + 142x - 120}{x - 1} \\ \\ = > (x - 12)(x - 10)$


$Therefore, \\ \\ {x}^{3}-23 {x}^{2} + 142x - 120 = (x - 1)(x - 12)(x - 10)$