How do you factor $10t^2 - 19t + 6$ ?

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Answer 1 · 2018-03-10T08:41:05

$(5t-2)(2t-3)$

$10t^2-19t+6$

1) multiply the coeff $" "x^2$ and the constant.

$color(red)(10)t^2-19t+color(red)(6)$

$10xx6=60$

2) find the factors of this number that add to the middle term ie. $" "-19$

$-15, -4$

3) now split the middle term with these factors and then factorise by grouping

. $10t^2 color(blue)(-19t)+6$

$10t^2color(blue)(-15t-4t)+6$

$(10t^2-15t)-(4t-6)$

$5t(2t-3)-2(2t-3)$

$(5t-2)(2t-3)$

How do you factor 10t^2 - 19t + 610t^2 - 19t + 6?