How do you factor $6 t^{2} + t - 1$ $6t^2+t-1$ ?

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Answer 1 · 2018-05-03T01:57:09

$(3 t - 1) (2 t + 1)$ $(3t - 1)(2t +1)$

Use the rainbow method.

Multiply the outer coefficients. $6 \cdot - 1 = - 6$ $6 * -1 = -6$

Now find two values which when multiplied equal the new number, and when added equal the center coefficient, 1.

Our values are 3 and -2.

Change the $+ t$ $+t$ to $+ 3 t$ $+3t$ and $- 2 t$ $-2t$

$6 t^{2} + 3 t - 2 t - 1$ $6t^2 + 3t - 2t - 1$

Separate the expression into two sides.
$(6 t^{2} + 3 t) and (- 2 t - 1)$ $(6t^2 + 3t) and (-2t - 1)$

Factor the two expressions, and make sure the values inside the parentheses are equal.

$3 t (2 t + 1) and - 1 (2 t + 1)$ $3t(2t + 1) and -1(2t +1)$

We now have $3 t - 1$ $3t - 1$ and $2 t + 1$ $2t +1$ as our factors.
$(3 t - 1) (2 t + 1)$ $(3t - 1)(2t +1)$

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