Question

Find the GCD of the following :
(i) 90, 150, 225
(ii) $15x^4{y}^3{z}^2,12x^2{y}^7{z}^2$
(iii) $6(2x^2-3x-2),8(4x^2+4x+1),12(2x^2+7x+3)$

Answer 1

(i) Let us write the numbers 90, 150 and 225 in the product of their prime factors as
$90\times =2\times 3\times 3\times 5,150=2\times 3\times 5\times 5$ and $225=3\times 3\times 5\times 5$
From the above 3 and 5 are common prime factors of all the given numbers.
Hence the GCD = $3\times 5=15$
(ii) We shall use similar technique to find the GCD of algebraic expressions. Now let us take the given expressions $15x^4{y}^3{z}^2$ and $12x^2{y}^7{z}^2$.
Here the common divisors of the given expressions are $3,x^2,y^3$ and $z^2$.
Therefore, GCD $=3\times x^2\times y^3\times z^2=3x^2{y}^3{z}^2$.
(iii) Given expressions are $6(2x^2-3x-2),8(4x^2+4x+1),12(2x^2+7x+3)$
Now, GCD of 6, 8, 12 is 2.
Next let us find the factors of quadratic expressions.
$2x^2-3x-2=(2x+1)(x-2)$
$4x^2+4x+1=(2x+1)(2x+1)$
$2x^2+7x+3=(2x+1)(x+3)$
Common factor of the above quadratic expressions is (2x+1).
Therefore, GCD = 2(2x+1).