How to factor #2x^2+3x+1=0# ?

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Answer 1 · 2017-12-11T21:42:06

#(2x+1)(x+1) = 0 #
#=> x = -1/2 , x= -1 #

To solve this we must understand how to factor:

if #y = ax^2 + bx + c #

Then you need to find two numbers who sum to make #b# but multiply to make #a*c#

So in this circumstance:

Plus to make #3# and multiply to make #2*1 = 2 #

Pairs of numbers who multiply to make #2#:

# ( 2 , 1) , (-2 ,-1 ) #

But out of these we see that #(2,1) # also plus to make #3#

So we can use the fact that #2x + x = 3x# :

#=> 2x^2 + 2x + x +1 #

Now we can factor the first 2 and last 2 terms individually, but ensuring they both have the same factor:

#=> 2x(x+1 ) + (x+1) #

Factoring out the #(x+1) # :

#=> (x+1)[2x + 1 ] #

#=> (x+1)(2x+1) #

If we want to solve for #0# then either one of #(x+1) or (2x+1) # must equal #0# and #alpha*0 = 0 #

So hence #(x+1) = 0 => x = -1 #

Also #(2x+1) = 0 => 2x = -1 => x = -1/2#