Q. 15.0( 1 Vote )

Let's find the G.C.D. of the algebraic expressions x(x2– 9), x2 – x – 12.

Let us understand what a G.C.D, Greatest Common Divisor 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.

To find G.C.D of x(x2 – 9) and x2 – x – 12, let us write down factors of each term.

Factorization of x(x2 – 9) = x(x – 3)(x + 3)

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

For factorization of x2 – x – 12,

x2 – x – 12 = x2 – 4x + 3x – 12 [, Sum of -4 and 3 is -1 and multiplication is -12]

x2 – x – 12 = x(x – 4) + 3(x – 4) [, common from the first two terms is x and last two terms is 3]

x2 – x – 12 = (x – 4)(x + 3) [, common from the two terms is (x – 4)]

So, factorization of x2 – x – 12 = (x – 4)(x + 3)

Find the factors that these two lists share in common.

Factors of x(x2 – 9) = x, (x – 3), (x + 3)

Factors of x2 – x – 12 = (x – 4), (x + 3)

Common factor that is found in these two terms is (x + 3).

Thus, the gcd of x(x2 – 9) and x2 – x – 12 is (x + 3).

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