How do you multiply $(3 t - 2) (4 t + 2)$ $(3t-2)(4t+2)$ ?

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Answer 1 · 2018-04-03T02:48:30

The answer is $12 t^{2} - 2 t - 4$ $12t^2-2t-4$ .

First,Multiply the equation $(4 t + 2)$ $( 4t +2)$ by $(3 t)$ $(3t)$ and $(4 t + 2)$ $( 4t +2)$ by $(- 2)$ $(-2)$ .

Converting it into an equation,

$\Rightarrow$ $rArr$ $3 t (4 t + 2) - 2 (4 t + 2)$ $3t(4t+2) -2(4t+2)$

$\Rightarrow$ $rArr$ $12 t^{2} + 6 t - 8 t - 4$ $12t^2+6t -8t-4$

$\Rightarrow$ $rArr$ $12 t^{2} + (6 - 8) t - 4$ $12t^2+(6-8)t-4$

$\Rightarrow$ $rArr$ $12 t^{2} - 2 t - 4$ $12t^2-2t-4$

$:.$ the answer is $12t^2-2t-4$ .

Let's hope this helps you:)

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