Multiply out and simplify to obtain a minimum sum-of-products boolean expression:
(A' + B +C')(A' +C' + D )(B' + D')

As per the question, we need to convert product of sum into sum of product,

Given:

(A' +B+C')(A'+C'+D)(B'+D'),

At first, we will solve to parenthesis,

= (A'+C'+BD) (B'+D')

As per the Rule, (A+B)(A+C) = A+BC, In our case if we assume X = A'+C', then,

(A' +B+C')(A'+C'+D) = (A'+C'+B)(A'+C'+D) = (A'+C'+BD)

Now,  

= (A'+C'+BD) (B'+D') = A'B' + A'D' + C'B' +C'D' +BDB' +BDD"

As we know that AA' =  0, it mean

=A'B'+A'D'+C'B'+C'D'+D*0+B0

=A'B'+A'D'+C'B'+C'D'               as B * 0 and D*0 = 0

Finally, minimum sum of product boolean expression is

A''B'+A'D'+C'B'+C'D'

=

Step-by-step explanation:

Standard form is

3n² + 17n - 6 = 0

Factored form is

(3n - 1)(n + 6) = 0