Question

X³–23x²+142x–120 using factor theoram

Answer 1

Step-by-step explanation:

Let p(x)= x³-23x²+142x-120

By hit and trial we find :

p(1) = (1)³-23(1)²+142(1)-120

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

p(1)=0

therefore (x-1) is a factor of p(x) (by factor theorem)

dividing p(x) by x-1

we get remainder is 0 and quotient as x²-22x+120

We know

Dividend = Divisor × quotient +Remainder

p(x) = (x-1) (x²-22x+120) +0

= x²- 12x- 10x+120

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

=(x-12)(x-10)

Hence x³-23x²+142x-120

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