Question

Using factor theorem, factorize each of the following polynomial:
x³-23x²+142x-120

Answer 1

The factorised given polynomial is $x^3-23x^2+142x-120=(x-1)(x-10)(x-12)$

Step-by-step explanation:

Given that the polynomial $x^3-23x^2+142x-120$

To factorise the given polynomial  :

Let f(x) be the given polynomial

$f(x)=x^3-23x^2+142x-120$

By using the Factor theorem here

• Put x=1 in the polynomial f(x) we get

$f(1)=1^3-23(1)^2+142(1)-120$

$=1-23+142-120$

$=143-143$

$=0$

$f(1)=0$

Therefore x-1 is a factor and it satisfies the given polynomial.( by factor theorem )                                  

• Put x=10 in f(x)  

$f(10)=(10)^3-23(10)^2+142(10)-120$

$=1000-23(100)+1420-120$

$=2420-2420$

$=0$

$f(10)=0$

Therefore x-10 is a factor and it satisfies the given polynomial.( by factor theorem )

• Put x=12 in f(x)

$f(12)=12^3-23(12)^2+142(12)-120$

$=1728-23(144)+1704-120$

$=3312-3312$

$=0$

$f(12)=0$

Therefore x-12 is a factor and it satisfies the given polynomial.( by factor theorem )

Therefore the factors are x-1,x-10,x-12

The factorised given polynomial is $x^3-23x^2+142x-120=(x-1)(x-10)(x-12)$

                                                                                           

Answer 2

Step-by-step explanation:

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