How do you factor $3 t^{2} - 8 t + 4$ $3t^2-8t+4$ ?

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How do you factor $3 t^{2} - 8 t + 4$ $3t^2-8t+4$ ?

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Answer 1 · 2017-04-22T15:40:09

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

Explanation:

3 is a prime number so we can only have: $(3 t \pm ?) (t \pm ?)$ $(3t+-?)(t+-?)$

The whole number factors of 4 are: 1 and 4 also 2 and 2
As the 4 is positive then both the ? must be of the same sign
As the 8 is negative it implies that both the ? are negative

Lets try:

$(3 t - 1) (t - 4) \to 3 t^{2} - 12 t - t + 4 Fail$ $(3t-1)(t-4) -> 3t^2-12t-t +4 color(red)(" Fail")$

Lets try

$(3 t - 2) (t - 2) \to 3 t^{2} - 8 t + 4 Works$ $(3t-2)(t-2)->3t^2-8t+4 color(red)(" Works")$

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How do you factor $3 t^{2} - 8 t + 4$ $3t^2-8t+4$ ?

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How do you factor $3 t^{2} - 8 t + 4$ $3t^2-8t+4$ ?