Question

x³-23x²+142x-120 factorise​

Answer 1

Answer:

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

→ 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)

Hope it helps...

Answer 2

Answer:

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

→ 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)

Hope it helps....…