Respuesta :

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]

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