Question

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Answer 1

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Answer 2

☺️hey mate! here's your answer!☺️


let p(x) = x³ - 23x² + 142x - 120

so, let us try using x = 1

then,

p(1) = 0

so, x - 1 is a factor of p(x)

then,

x³ - 23x² + 142x - 120

= x³ - x² - 22x² + 22x + 120x - 120

= x² ( x - 1 ) - 22x ( x - 1 ) + 120 ( x - 1 )

then, we observe that ( x - 1 ) is common!

= ( x - 1 ) ( x² - 22x + 120 )

as, we have got that (x - 1) is a factor.

also, the next term that is (x² - 22x + 120 ) can be done through middle term splitting method,

x² - 22x + 120

= x² + (-12 -10)x + 120

= x² - 12x - 10x + 120

so, we will take common,

= x ( x - 12) - 10(x - 12)

= (x - 12) ( x - 10)

thus other terms that are factor of p(x) are (x - 12) ( x - 10).

hence, x³ - 23x² + 142x - 120 = ( x - 1 ) (x - 12) ( x - 10).

hope it is helpful ✌️