Find the zeroes of the quadratic polynomial 6x² - 13x +6 and verify the relation between the zeroes and its coefficients.

We have,
[tex]6x {}^{2} - 13x + 6 = 6x {}^{2} - 4x - 9x + 6 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2x(3x - 2) - 3(3x - 2) = (3x - 2)(2x - 2)[/tex]

So, to find zeroes of polynomial : 6x² - 13x +6 will be 0,hence ( 3x-2)= 0 and (2x-3) = 0.

So, [tex]x = \frac{2}{3} \: \: and \: x = \frac{3}{2} [/tex]

Therefore, the zeroes of 6x² - 13x +6 are [tex] \frac{2}{3 } \: and \: \frac{3}{2} [/tex]

Sum of the zeroes
[tex] = \frac{2}{3} + \frac{3}{2} = \frac{13}{6} = \frac{ - ( - 13)}{2} = \frac{ - coefficient \: of \: x}{coefficient \: of \: x {}^{2} } [/tex]

Product of zeroes
[tex] = \frac{2}{3} \times \frac{3}{2} = \frac{6}{6} = \frac{constant \: term}{coefficient \: of \: x {}^{2} } [/tex]

Hence verified.