Question

Factorise x²- 23x²+ 142x-120​

Answer 1

Step-by-step explanation:

We know that if the sum of the coefficients 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

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)