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

Answer 1

Eli Assouline

#(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#

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Answer 2

Eva Abramson

To factor the quadratic equation (2x^2 + 3x + 1 = 0), you can use the quadratic formula (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 2), (b = 3), and (c = 1). Then, solve for (x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.