the GCD of 2002,k=4,then what is the value of k
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x2 − 2x − 15
= x2 − 5x + 3x − 15 [− 5x + 3x = −2x, − 5x × 3x = −15x2]
= x(x − 5) + 3(x − 5)
= (x − 5) (x + 3)
x3 + 27
= (x)3 + (3)3
= (x + 3) (x2 − 3x + 9) [a3 + b3 = (a + b) (a2 − ab + b2)]
∴GCD of x2 − 2x − 15 and x3 + 27 is x + 3.
It is given that x + k is the G.C.D of x2 − 2x − 15 and x2 + 27.
Comparing x + 3 with x + k, we get k = 3.
Thus, the value of k is 3.
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