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I WILL GIVE BRAINLIEST IF YOU FIGURE IT OUT

Give a third degree polynomial that has zeros of 17, 7i, and −7i, and has a value of −845 when x=4. Write the polynomial in standard form axn+bxn−1+….

Respuesta :

angstar4
p(x) = x^3 - 17x^2 + 49x - 833

p(x) = (x-17)(x^2+49)
7i * 7i = -49

p(x) = x^3 - 17x^2 + 49x - 833
p(4) = 4^3 - 17*4^2 + 49*4 - 833
p(4) = 64 - 17*16 + 196 - 833
p(4) = 64 - 272 - 637
p(4) = -208 - 637
p(4) = -845 :)

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