Find the zeros of polynomial f (x) =x^3-5x^2-16x+80, it its two zeros are equal is magnitude but opposite in sign.
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F(x) = X^3 - 5x^2 -16x +80
Let its zeros be P, Q,-P (as its 2 zeroes are eq in magnitude but opposite signs)
Sum of zeroes = -b/a
P+(-P)+Q = -(-5)/1
Q = 5........(1)
Now, Product of zeroes = -d/a
P(-P)(B) = -80/1
-p^2 x Q = -80
Using (1)
P^2 = -80/5
p x p = 4 x 4
P= 4
Hence zeroes are 5, 4, -4
Hope it will help you ...And Brainliest plz...
Let its zeros be P, Q,-P (as its 2 zeroes are eq in magnitude but opposite signs)
Sum of zeroes = -b/a
P+(-P)+Q = -(-5)/1
Q = 5........(1)
Now, Product of zeroes = -d/a
P(-P)(B) = -80/1
-p^2 x Q = -80
Using (1)
P^2 = -80/5
p x p = 4 x 4
P= 4
Hence zeroes are 5, 4, -4
Hope it will help you ...And Brainliest plz...
srushtikadam4:
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