Thread: factor the equation

3q^2 + 16q + 5

I look for a solution of the form (3q-a)(q-b). Expanding . Equating the coefficients of like terms gives ab=5 and 3b+a=-16. Thinking for a moment will convince you that b=-5 and a=-1 satisfy these equations. So .

No......my previous answer is wrong......sorry about tht!!!

3q^2 + 16q +5

3q^2 +15q +q +5

3q( q+5) +1 (q+5)

(3q+1) (q+5)

Originally Posted by kenzie103109

3q^2 + 16q + 5

The rational roots theorem suggests that , and are the only possible rational roots for this quadratic in . Descartes rule of signs tells you it has no positive roots, so that reduces the list to and . Now trying these out shows that and , so we have:



and to match the leading coefficient of we must have so:



CB