Answer:
[tex]y^2-\frac{7}{2}y+3[/tex]
Step-by-step explanation:
Given polynomial is,
6y² - 7y - 2
Fro the zeros of the given polynomial,
6y² - 7y - 2 = 0
6y² - 4y - 3y - 2 = 0
2y(3y - 2) - 1(3y - 2) = 0
(2y - 1)(3y - 2) = 0
(2y - 1) = 0
y = [tex]\frac{1}{2}[/tex]
(3y - 2) = 0
y = [tex]\frac{2}{3}[/tex]
Therefore, zeros of this polynomial are m = [tex]\frac{1}{2}[/tex] and n = [tex]\frac{2}{3}[/tex]
If a polynomial has zeros as [tex]\frac{1}{m}[/tex] and [tex]\frac{1}{n}[/tex] then the zeros will be [tex]\frac{3}{2}[/tex] and 2.
Polynomial will be,
[tex](y-\frac{3}{2})(y-2)[/tex]
= [tex]y(y-2)-\frac{3}{2}(y-2)[/tex]
= [tex]y^2-2y-\frac{3}{2}y+3[/tex]
= [tex]y^2-\frac{7}{2}y+3[/tex]