Math, asked by jagusoni7777pe4t8e, 10 months ago

factorise x^3-23x^2+142x-120​

Answers

Answered by rishita4678
6

x3-23x2+142x-120

by trial and error method let us see if [x-1] is a factor of this polynimial

therefore p[1]= 1³-23×1²+142×1-120

                    = 1-23+142-120

                    = 0

so [x-1] is a factor,

now, to divide [x-1] with  x3-23x2+142x-120

by doing this we will get x²-22x+120

to find the other factors of  x3-23x2+142x-120 we can split the middle term

x3-23x2+142x-120

=x²-12x+[-10x] +120

=x[x-12]-10[x-12]

=[x-10][x-12]

therefore the factors of  x3-23x2+142x-120 are:

[x-1][x-10][x-12]

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