Question

factorise x^3-23x^2+142x-120​

Answer 1

Answer:

5+8(=7+;#)#£4(#)973148/!)':+#:(_[`°<°%©«77

Answer 2

$\tt\LARGE{\red{Answer:-}}$

(x-1)(x+12(x+10)

$\tt\LARGE{\orange{Explanation:-}}$

Let p(x)= x3- 23x2+142x-120

g(x)=x-1

0=x-1

x=1

p(1)=(1)3-23(1)2+142(1)-120

=1-23+142-120

=143-143

=0

Hence x-1 is a factor of x3- 23x2+142x-120

By long division

x3- 23x2+142x-120 = (x-1)(x2+22x+120)

=(x-1)(x2+10x+12x+120)

=(x-1)(x(x+10) 12(x+10))

=(x-1)(x+12)(x+10)

Hope it helps u...✌️❣️