Question

One of the factors of p(x) = x3 – 23x2 + 142x – 120 is
(x + 1)
(x + 10
(x-8)
(x-12)​

Answer 1

Answer:

x-12

Step-by-step explanation:

thats it.

Answer 2

Step-by-step explanation:

Given that: One of the factors of p(x) = x³ – 23x² + 142x – 120 is

(x + 1)

(x + 10

(x-8)

(x-12)

Solution:

To find the factor out of the given options

Put the factor equal to zero,if on putting that value p(x)=0,

then that particular option is correct.

$p(x) = {x}^{3} - 23 {x}^{2} + 142x - 120$

Put

$x + 1 = 0 \\ \\ x = - 1$

in the polynomial

$p( - 1) = {( - 1)}^{3} - 23 {( - 1)}^{2} + 142( - 1) - 120 \\ \\ p( - 1)= - 1 - 23 - 142 - 120 \\ \\ p( - 1) = - 286 \\ \\ p( - 1) \neq0$

so, it can't be a factor.

(x+10) also not have a factor of p(x) because on putting x= -10,all quantity becomes negative,nit equal to zero

try for x-12

x= 12

$p( 12) = {(12)}^{3} - 23( {12)}^{2} + 142 \times 12 - 120 \\ \\ = 1728 - 3312 + 1704 - 120 \\ \\ = 3432 - 3432 \\ \\ p(12) = 0$

Thus,

(x-12) is a factor of p(x).

Option 4 is correct.

Hope it helps you.