can you factor 2x^2+3x+1=0?

Question

i'm brain dead ):

& .. 3u^2= -4u + 15

25v^2 -30v = -19

Answer 1

1. (2x+1)(x+1)
x= -1/2 x= -1
2. (3u-5)(u+3)
u=5/3 u= -3
3.
25v^2 - 30 +19=0
a=25
b= -30
c= 19
[-b +/-√(b^2 - 4ac)]/2a
[-30 +/- √(900 -1900)]50
[-30 +/- √-1000]/50
[-30 +/- 10i√10]/50
(-3 +/- i√10)/5

Answer 2

yup
factors of 2 (1,2)
therefore
(2x+1)(x+1)

next
3u^2= -4u + 15
have to get everybody on the same page
add 4u to both sides
3u^2 +4u = 15...now subtract 15 both sides
3u^2 +4u -15= O
3 factors (1,3) and 15 factors(1,3,5,15)
so 1 and 3 are givens, my bet is the inner factors for 15
and since the middle is a + then the larger combo is the positive
(3u - 5)(u + 3)

same thing goes for # 3 - but it's trickier
NOPE - recheck your equation - something is WRONG
- all the best

Answer 3

2x^2+3x+1=0
=> 2x^2 + 2x+ x + 1
=> 2x ( x + 1 ) + 1 ( x + 1 )
=> ( 2x + 1 ) ( x + 1 ) ................ Answer

........................................................................

3u^2= -4u + 15
=> 3u^2 + 4u -15 = 0
=> 3u^2 + 9u - 5u - 15 = 0
=> 3u ( u + 3 ) - 5 ( u + 3 ) = 0
=> ( 3u - 5 ) ( u + 3 ) = 0
=> Either ( 3u - 5 ) = 0 so that u = 5/3 ............. Answer
Or ( u + 3 ) = 0 => u = - 3 .............. Answer
Hence soln is x : [ 5/3, -3 ]

25v^2 -30v = -19
This eqn has imaginary roots but can be solved.

Answer 4

2x^2+3x+1=0
(2x +1)(x +1) = 0

3u^2 = -4u +15
3u^2 + 4u - 15 = 0
(3u - 5)(u +3)

hope this helps :-)