Factorise $6 t^{2} + 17 t + 5$ .

(2 t + 1) (3 t + 5)

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(2 t + 5) (3 t + 1)

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(6 t + 5) (t + 1)

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(2 t + 3) (3 t + 2)

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Solution

The correct option is B $(2 t + 5) (3 t + 1)$
Factorising above equation ;

$\Rightarrow 6 t^{2} + 15 t + 2 t + 5$

$\Rightarrow 3 t (2 t + 5) + 1 (2 t + 5)$

$\Rightarrow (2 t + 5) (3 t + 1)$

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Similar questions

f (t) = \frac{4}{t} - \frac{1}{6 t^{3}} + \frac{8}{t^{5}}

\frac{9}{10} (30 + 5 t) + \frac{5}{6} (18 t - 12)

\frac{3 t - 2}{4} - \frac{2 t + 3}{3} = \frac{2}{3} - t

t-(2t+5)-(1-2t)=(3+4t)-2(t-4)

With checking

2 t^{2} + 3 t^{3} - 2 t + 5

t = 1

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