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divide by 3: 3(6x2 + 13x - 5)
Look for factors of 6 which can be multiplied by factors of -5 and added to give 13.
Factors of 6: 1 & 6; 2 & 3;
Factors of -5: -1 & 5; -5 & 1
the last and first ones give 3 x -5 = -15 and 2 x 1 = 2, total 13 so:
3(3x - 1)(2x + 5) = 3(6x2 - 2x + 15x - 5) = 18x2 + 39x - 15
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How do you factor the trinomial completely 18x3 29x 3x?
I'm guessing that's 18x3 + 29x2 + 3x. That factors to x(2x + 3)(9x + 1) 18x3 + 29x2 + 3x factor x = x(18x2 + 29x + 3) write 29x as 27x + 2x = x(18x2 + 27x + 2x + 3) group by two and factor each group = x[(18x2 + 27x) + (2x + 3)] = x[9x(2x + 3) + 1(2x + 3)] continue factoring = x(2x + 3)(9x + 1) Or since (18)(3) = 54 and (27)(2) = 54 then write: = x(18x2 + 29x + 3) = x[(18x + 27)/9][(18x + 2)/2] = x(2x + 3)(9x + 1)
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How do you factor the trinomial completely 18x3 29x 3x?
I'm guessing that's 18x3 + 29x2 + 3x. That factors to x(2x + 3)(9x + 1) 18x3 + 29x2 + 3x factor x = x(18x2 + 29x + 3) write 29x as 27x + 2x = x(18x2 + 27x + 2x + 3) group by two and factor each group = x[(18x2 + 27x) + (2x + 3)] = x[9x(2x + 3) + 1(2x + 3)] continue factoring = x(2x + 3)(9x + 1) Or since (18)(3) = 54 and (27)(2) = 54 then write: = x(18x2 + 29x + 3) = x[(18x + 27)/9][(18x + 2)/2] = x(2x + 3)(9x + 1)
