How do you factor #6t^2+t-1#?

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Answer 1 · 2018-05-03T01:57:09

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

Use the rainbow method.

Multiply the outer coefficients. #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# to #+3t# and #-2t#

#6t^2 + 3t - 2t - 1#

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

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

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

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