Question

find all the zeroes p(x)= x^3-8x^2+19x-12, having one of its zeroes as 4​

Answer 1

Answer:alpha = 3, beta = 1

Step-by-step explanation:

It is given that 4 is the zero of px, then (x-4)

is completely divisible by px

So we divide px by (x-4)

Quotient = x^2 -4x +3

By factorizing further,

= (x-1)(x-3)

So, the zeroes are 1 and 3

Hope it helps