Problem 1:
=y3 + y2-4y-4
In order to solve this equation, we first take out common factors i.e for (y3 + y2), y2 is a common factor and for (-4y-4), -4 is a common factor
=y2(y+1)-4(y+1)
Combining common factors and arranging the brackets, we get;
=(y2-4)(y+1)
Equating both factors with 0, we get:
(y2-4)(y+1)=0
Y= -1, 2 and -2
Hence, value of Y= -1, 2 and -2
Problem 2:
=t2+ 4z2
Since the equation have two variables i.e t and z, we really cannot have real number answer to this equation. However, we can arrange perfect squares of these numbers
=(t)2+ (2z)2
Hence, this equation will stop over here only without being factored.
Problem 3:
=9t2 + 5t- 4
Using AC method to break the middle term i.e 5t so that the resultant sum is 5t, while multiplying the factors we get 9t2*-4= -36t2. Hence breaking 5t in such a way we get 9t and -4t such that resultant sum is 5t and multiplying these two terms we indeed get -36t2
= 9t2+ 9t-4t-4
In order to solve this equation, we first take out common factors i.e for (9t2 + 9t), 9t is a common factor and for (-4t-4), -4 is a common factor. Thus arranging the factors we get;
=9t(t+1)-4(t+1)
Combining common factors and arranging the brackets, we get;
=(9t-4)(t+1)
Equating both factors with 0, we get:
(9t-4)(t+1)= 0
t= -1, 4/9
Hence, value of t= -1, 4/9
