trinomials?

Question

Summer school homework. and i don't understand these kind of problems. Can someone please help me out. <3 thanks. its much appriciated.
Problem:
Factor the trinomial.
5x^2-5x-3

Answer 1

lol...algebra 1 i forgot all of that my after 8th grade.

Answer 2

factoring....(if it is factorable)

[after posting... this equation is NOT factorable...
the -5 is either a -2, or, if it is -5, then need to use the
quadratic equation as other have done above]


5x^2 - 5x - 3 = (ax+b)(cx+d)

where a, b, c and d tend to be integers.
If you foil this out....

ac(x^2) + (ad+bc)x +bd = 5x^2 - 5x - 3

a*c = 5
ad + bc = -5
b*d = -3

5 is a prime, so likely either a = +/-5, +/-1, or c = +/-5, +/-1
3 is a prime, so likely either b = +/-3, +/-1, or d = -+3, -/+1
ad + bc also must be negative.
You can guess and check.

Can substitute different sets of possible a,b,c, d and check the foil...
a*c = 5
b*d = -3
ad + bc = -5

a c b d
1 5 -3 1 .....
ac = 5 OK
bd = -3 OK
ad + bc = -13 NO

a c b d
1 5 3 -1 .....
ac = 5 OK
bd = -3 OK
ad + bc = 13 NO

a c b d
5 1 -3 1 .....
ac = 5 OK
bd = -3 OK
ad + bc = 2 NO

a c b d
5 1 3 -1 .....
ac = 5 OK
bd = -3 OK
ad + bc = -2 NO


?????
is the equation write (or is it too late for me).

a = 5, c=1, b=3, d=-1

(5x+3) (x-1) .... check,,, FOIL: 5x^2 -5x + 3x -3 = 5x^2 - 2x -3

Answer 3

Given an equation of the format ax^2 + bx + c = 0

x can be calculated as

x = (-b +/- sqrt(b^2 * 4 * a * c)) / (2 * a)

Here:

a = 5
b = -5
c = -3

x = (5 +/- sqrt((-5)^2 - 4 * 5 * -3)) / (2 * 5)

x = (5 +/- sqrt(25 + 60)) / 10

x = 5/10 +/- sqrt(85) / 10

x = .5 +/- sqrt(85) / 10

sqrt(85) = 9.22

x = .5 + .92 or .5 - .92

x = 1.42 or -.42

Answer 4

if u use the quadratic formula u get x= 1.4219 and x= -.4219.
or 5 + or - the square root of 85 all over ten.
i think =]

Answer 5

no solution...i've tried everything...but if there is an answer then let me know.