Question

Factorise :

x³ - 23x² + 142x - 120

Quality answer needed ... No wrong answers .​

Answer 1

Question:

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

Solution:

$\implies \: {x}^{3} - {x}^{2} - {22x}^{2} + 22x + 120x - 120$

$\implies \: {x}^{2} (x + 1) - 22x(x + - 1) + 120(x + 1)$

$\implies \: (x - 1)( {x}^{2} - 22x + 120)$

$\implies \: (x - 1)( {x}^{2} - 10x - 12x + 120)$

$\implies \: (x - 1)(x(x - 10) - 12(x - 10))$

$\boxed{\implies \: (x - 1)(x - 10)(x - 12)}$

Answer 2

Answer:

we know that if the sum of the coefficients is equal to 0 then (x-1)is one of the factor of given polynomial

x-1 = 0

x=1

put x=1

(1)³-23(1)²+142(1)-120

➡️1-23+142-120

➡️120-120

➡️0

therefore, (x-1) is a factor of the given polynomial.

now the factors are (x-1) and (x²-22x+120)

now factories x²-22x+120

x²-22x+120

➡️x²-12x-10x+120

➡️x(x-12)-10(x-12)

➡️(x-12)(x-10)

therefore, x²-22x+120 = (x-12)(x-10)

the factors of x³-23x²+142x-120

➡️(x-1)(x-12)(x-10)