find the zeros of y3-3y2+4y+2
Answers
Answered by
7
A little mistake is happened in question .
A question may be ...... y³ - 3y² + 4y - 2
= y³ - y² - 2y² + 2y + 2y - 2
= y²(y - 1) - 2y(y - 1) + 2(y - 1)
= (y² - 2y + 2)(y - 1)
Hence, y = 1 is the real zero of given polynomial.
now, y² - 2y + 2
Check discriminant , D = b² - 4aC
D = (-2)² - 4(2) = - 4 < 0
discriminant is negative ∴ zeros are imaginary numbers.
e.g., y = (2 ± 2i)/2 = (1 ± i)
Hence, y = 1, 1 + i and 1 - i are the zeros of polynomial.
A question may be ...... y³ - 3y² + 4y - 2
= y³ - y² - 2y² + 2y + 2y - 2
= y²(y - 1) - 2y(y - 1) + 2(y - 1)
= (y² - 2y + 2)(y - 1)
Hence, y = 1 is the real zero of given polynomial.
now, y² - 2y + 2
Check discriminant , D = b² - 4aC
D = (-2)² - 4(2) = - 4 < 0
discriminant is negative ∴ zeros are imaginary numbers.
e.g., y = (2 ± 2i)/2 = (1 ± i)
Hence, y = 1, 1 + i and 1 - i are the zeros of polynomial.
Similar questions