Factorize $6 t^{2} + 13 t + 7$ and verify the relationships between the zeroes and the coefficients.

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Solution

$6 t^{2} + 13 t + 7$

= $6 t^{2} + 6 t + 7 t + 7$

= (t+1)(6t+7)

So the zeroes are -1, $\frac{- 7}{6}$

$α + β$ = -1 + $\frac{- 7}{6}$ = $\frac{- 13}{6}$ = $\frac{- b}{a}$

$α β$ = $\frac{7}{6}$ = $\frac{c}{a}$

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