Question

Factorise :

x³ - 23x² + 142x - 120

Quality answer needed ( No wrong answers .)​

Answer 1

●We know that the sum of thecoefficients is equal to 0 then (x-1) is one of the factors 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) {see pic for knowing how (x²-22x+120) is a factor}

Now factorise 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)

Answer 2

Step-by-step explanation:

${x}^{3} - {x}^{2} - 22 {x}^{2} + 22x + 120x + 120$

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

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

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

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

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