Question

5s^2+19s-4 help please

Answer 1

what is the question? factor?

Answer 2

yes factor

Answer 3

abc formula: -b/2a + or - sqrt (b^2-4ac)/2a

Answer 4

(a+4)(5a-1)

Answer 5

sorry in place of a u can write s

Answer 6

find non-zero integers a b c d such that \[(a*s + b)(c*s + d) = 5s^2 + 19s - 4\]and \[a * c = 5\]\[b * d = -4\]\[b*c + a*d = 19\]
5 is prime so we've found \[a = 5\ \ \ and \ \ \ c = 1\]
hence our problem is now to find b and d such that \[(5s + b)(s + d) = 5s^2 + 19s - 4\]where\[b*d = -4\]and \[b + 5d = 19\]

it follows from \[b+5d = 19\]that \[b = 19 - 5d\]hence \[19d - 5d^2 = -4\]\[5d^2 - 19d -4 = 0\]solving gives us \[d = -\frac{1}{5} \ \ \ or \ \ \ d = 4\]we pick the integer solution
hence \[b = -1\]
therefore \[a = 5\]\[b = -1\]\[c = 1\]\[d = 4\]


hence \[5s^2 +19s - 4 = (5s - 1)(s + 4)\]
.... should have used that solution formula in the first step.