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Question
If the zeros of the polynomial f(x) = x3 − 12x2 + 39x + k are in A.P., find the value of k.
Solution
Let a - d, a and a + d be the zeros of the polynomial f(x). Then,
Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`
`a-d+a+a+d=(-(-12))/1`
`a-d+a+a+d=12`
`3a=12`
`a=12/3=4`
Since 'a' is a zero of the polynomial f(x).
f(x) = x3 − 12x2 + 39x + k
f(a) = 0
f(a) = 43 − 12(4)2 + 39(4) + k
0 = 64 - 192 + 156 + k
0 = 220 - 192 + k
0 = 28 + k
-28 = k
Hence, the value of k is -28.
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