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  • Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials: 7y^2

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

7y^{2}-\frac{11}{3}y-\frac{2}{3}

Answers (1)

Answer. \left [\frac{2}{3},\frac{-1}{7} \right ]

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

7y^{2}-\frac{11}{3}y-\frac{2}{3}=0

Multiply by 3

21y^2 - 11y -2=0

21y^2 - 14y + 3y - 2=0

7y(3y -2) + 1 (3y - 2)=0

(3y - 2) (7y + 1)=0

3y - 2 = 0                                         7y + 1 = 0

y = \frac{2}{3}                                                   y = \frac{-1}{7}

Hence, \left [\frac{2}{3},\frac{-1}{7} \right ]  are the zeroes of the polynomial

Here, a = 21, b = –11, c = –2

Sum of zeroes \frac{-b}{a}=\frac{-(-11)}{21}=\frac{11}{21}

Here, \frac{2}{3}-\frac{1}{7}=\frac{14-3}{21}=\frac{11}{21}=\frac{-b}{a}

Product of zeroes \frac{c}{a}=\frac{-2}{21}

Here, \frac{2}{3}\times \frac{-1}{7}=\frac{-2}{21}=\frac{c}{a}

 

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