Find a polynomial P(x) of degree 3 that has zeros 2 and 3-i.
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Step-by-step explanation:
Solution:-
This is the way to find the answer
f(x) = (x-3)[x - (2+i)][x - (2-i)]
= (x-3)[(x-2) - i][(x-2) + i]
= (x-3)[(x-2)2 - i2]
= (x-3)(x2 - 4x + 5)
= x3 - 4x2 + 5x - 3x2 + 12x - 15 = x3 - 7x2 + 17x - 15
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Step-by-step explanation:
ANSWER ✍️
f(x) = (x-3)[x - (2+i)][x - (2-i)]
= (x-3)[(x-2) - i][(x-2) + i]
= (x-3)[(x-2)2 - i2]
= (x-3)(x2 - 4x + 5)
= x3 - 4x2 + 5x - 3x2 + 12x - 15 = x3 - 7x2 + 17x - 15
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