Q. 1

Let's find G.C.D. and L.C.M. of (y3 – 8), (y3 – 4y2 + 4y) and (y2 + y – 6)

Answer :

Let us understand what a G.C.D, Greatest Common Divisor and L.C.M, Least Common Multiple is.

The greatest common divisor of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.


Least common multiple of two or more integers, is the smallest positive integer that is divisible by these two or more integers.


To find G.C.D of (y3 – 8), (y3 – 4y2 + 4y) and (y2 + y – 6), let us write down factors of each term.


For factorization of y3 – 8,


Factorization of y3 – 8 = Factorization of y3 – 23


[, y3 – 8 = y3 – 23, as 23 = 2 × 2 × 2 = 8]


y3 – 8 = (y – 2)(y2 + 22 + (2)(y))


y3 – 8 = (y – 2)(y2 + 4 + 2y)


y3 – 8 = (y – 2)(y2 + 2y + 4)


[, by algebraic identity, a3 – b3 = (a – b)(a2 + b2 + ab)]


For factorization of y3 – 4y2 + 4y,


y3 – 4y2 + 4y = y(y2 – 4y + 4)


y3 – 4y2 + 4y = y(y2 – (2y + 2y) + 4) [, Sum of -2 and -2 is -4 and multiplication is 4]


y3 – 4y2 + 4y = y(y2 – 2y – 2y + 4)


y3 – 4y2 + 4y = y(y(y – 2) – 2(y – 2)) [, common from the first two terms is y and last two terms is -2]


y3 – 4y2 + 4y = y((y – 2)(y – 2)) [, common from the two terms is (y – 2)]


y3 – 4y2 + 4y = y(y – 2)(y – 2)


y3 – 4y2 + 4y = y(y – 2)2


For factorization of y2 + y – 6,


y2 + y – 6 = y2 + 3y – 2y – 6 [, Sum of 3 and -2 is 1 and multiplication is -6]


y2 + y – 6 = y(y + 3) – 2(y + 3) [, common from the first two terms is y and last two terms is -2]


y2 + y – 6 = (y + 3)(y – 2) [, common from the two terms is (y + 3)]


Find the factors that these two lists share in common.


Factors of y3 – 8 = (y – 2), (y2 + 2y + 4)


Factors of y3 – 4y2 + 4y = y, (y – 2), (y – 2)


Factors of y2 + y – 6 = (y + 3), (y – 2)


Common factor that is found in these three terms is (y – 2).


So, gcd is (y – 2).


To find L.C.D of (y3 – 8), (y3 – 4y2 + 4y) and (y2 + y – 6), let us write down factors of each term.


Then, we need to find the factor of highest power.


Factors of y3 – 8 = (y – 2), (y2 + 2y + 4)


Factors of y3 – 4y2 + 4y = y, (y – 2)2


Factors of y2 + y – 6 = (y + 3), (y – 2)


The factors are y, (y – 2)2, (y + 3) and (y2 + 2y + 4).


Multiplying these factors, we get


y × (y – 2)2 × (y + 3) × (y2 + 2y + 4) = y(y – 2)2(y + 3)(y2 + 2y + 4)


So, lcm is y(y – 2)2(y + 3)(y2 + 2y + 4).


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