How do you multiply $(2 t + 3) (3 t - 5)$ $(2t + 3)(3t - 5)$ ?

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How do you multiply $(2 t + 3) (3 t - 5)$ $(2t + 3)(3t - 5)$ ?

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Answer 1 · 2018-05-18T09:09:09

$6 t^{2} - t - 15$ $6t^2-t-15$

Explanation:

$each term in the second bracket is multiplied by each$ $"each term in the second bracket is multiplied by each"$
$term in the first bracket$ $"term in the first bracket"$

$\Rightarrow (2 t + 3) (3 t - 5)$ $rArr(color(red)(2t+3))(3t-5)$

$= 2 t (3 t - 5) + 3 (3 t - 5) \leftarrow distribute$ $=color(red)(2t)(3t-5)color(red)(+3)(3t-5)larrcolor(blue)"distribute"$

$= (2 t \times 3 t) + (2 t \times - 5) + (3 \times 3 t) + (3 \times - 5)$ $=(color(red)(2t)xx3t)+(color(red)(2t)xx-5)+(color(red)(3)xx3t)+(color(red)(3)xx-5)$

$= 6 t^{2} - 10 t + 9 t - 15 \leftarrow collect like terms$ $=6t^2color(magenta)(-10t)color(magenta)(+9t)-15larrcolor(blue)"collect like terms"$

$= 6 t^{2} - t - 15$ $=6t^2-t-15$

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How do you multiply $(2 t + 3) (3 t - 5)$ $(2t + 3)(3t - 5)$ ?

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