Question

Factorise x3-23x2+142x-120

Answer 1

[tex] x^{3}- x^{2} -22 x^{2} +22x+120x+120 [/tex]

[tex] x^{2} (x-1)-22x(x-1)+120(x-1) [/tex]

$(x-1) ( x^{2} -22x+120)$

$(x-1) ( x^{2} -12x-10x+120)$

$(x-1) (x(x-12)-10(x-12))$

$(x-1) (x-10) (x-12)$

Answer 2

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]