(ax-7)(bx+5) = 6x2+cx-35 If ab=6 and a-b=-1 what are two possible values of c?
I don't know if I'm on the right track, but so far I've found out that c=5a-7b
Please solve this problem for me and show me how you did it! Thanks :)
a) c=2 or c=-4
b) c=11 or c=-11
c) c=-2 or c=4
d) c=12 or c=-5
+129852 (ax-7)(bx+5) = 6x^2+cx-35 If ab = 6 and a - b = -1 what are two possible values of c?
(ab x^2 - 7bx + 5ax - 35) = 6x^2 + cx -35
6x^2 + ( 5a - 7b) x - 35 = 6x^2 + cx - 35
5a - 7b = c
If a - b = - 1 then b = a + 1
And
ab = 6 ⇒ b = 6/a
So
b = b
a + 1 = 6/a multiply through by a and rearrange
a^2 + a - 6 = 0 factor
(a - 2) ( a + 3) = 0
Set each factor to 0 and solve and we get that a = 2 , a = -3
When a = 2 b = 6 / a = 3
When a = - 3 b = 6/-3 = -2
So
(a, b ) = (2, 3) or ( -3, -2)
Using 5a - 7b = c
5(2) - 7(3) = c = - 11
5(-3) - 7(-2) = c = -1
Proof
a = 2 b = 3
( 2x - 7) ( 3x + 5) = 6x^2 - 21x + 10x - 35 = 6x^2 - 11x - 35 ⇒ c = 11
a = - 3, b = -2
(-3x - 7) (-2x + 5) = 6x^2 - 15x + 14x - 25 = 6x^2 - 1x - 35 ⇒ c = -1
