Question

factorise x^3 - 23x^2 +142x - 120

Answer 1

hey dear


here is your answer


Solution


Let P (x) x^3 - 23x^2 +142x - 120


firstly we assumed p(x) =0

p (0) = x ( 0) ^3 - 23 ( 0) ^2 + 142 ( 0) - 120

= 0 - 120

= - 120

it is not verified


secondly we assumed

p ( 1) = x ( 1) ^3. -23 ( 1) ^2 + 142 ( 1) -120

= 1 - 23 +142 -120

= 143 - 143

= 0

hence it's verified


So the factor is x + 1 = 0
x = -1


hence your answer is

( x - 1) it is a factor


hope it helps

thank you

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....…