SOLUTION: Solve by factoring (please show work)
{{{ x^2+5x+6=0 }}}
and
{{{ 2x^2-x-3=0 }}}
If you can help with both that would be much appreciated, but if you can only solve one
Algebra ->
Logarithm Solvers, Trainers and Word Problems
-> SOLUTION: Solve by factoring (please show work)
{{{ x^2+5x+6=0 }}}
and
{{{ 2x^2-x-3=0 }}}
If you can help with both that would be much appreciated, but if you can only solve one
Log On
Question 505293: Solve by factoring (please show work)
and
If you can help with both that would be much appreciated, but if you can only solve one i understand. Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website!
x^2+5x+6=0
x^2+3x+2x+6=0
x(x+3)+2(x+3)=0
(x+2)(x+3)=0
x+2=0 or x+3=0
x=-2 or x=-3
Check
Put x=-2 in original equation
x^2+5x+6=0
(-2)^2+5(-2)+6=0
4-10+6=0
-6+6=0
0=0
2x^2-x-3=0
2x^2-3x+2x-3=0
x(2x-3)+1(2x-3)
(2x-3)(x+1)=0
2x-3=0 or x+1=0
2x=3 or x=-10
x=3/2 or x=-1
Check
2x^2-x-3=0
Put x=3/2 in original equation
2(3/2)^2-(3/2)-3=0
2(9/4)-(3/2)-3=0
(9/2)-(3/2)-3=0
Multiply by 2 both sides of above equation
2*(9/2)-2*(3/2)-2*3=0
9-3-6=0
6-6=0
0=0
