find zeroes of p(x) x^-3x-18=o
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Hii There!!!
Here is your Answer
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=> Given, p(x) = x^2-3x-18=0
=> x^2 - 6x + 3x - 18 = 0
=> x ( x - 6 ) + 3 ( x - 6) = 0
=> ( x - 6 ) ( x + 3 ) = 0
=> x - 6 = 0 or x + 3 = 0
=> x = 6 or x = -3
~~~$o the Zeroes of p(x) are 6 and - 3 respectively
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$miley
¢heeku
# Dreamer
====================
Here is your Answer
____________________
=> Given, p(x) = x^2-3x-18=0
=> x^2 - 6x + 3x - 18 = 0
=> x ( x - 6 ) + 3 ( x - 6) = 0
=> ( x - 6 ) ( x + 3 ) = 0
=> x - 6 = 0 or x + 3 = 0
=> x = 6 or x = -3
~~~$o the Zeroes of p(x) are 6 and - 3 respectively
=====================
$miley
¢heeku
# Dreamer
====================
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