How do you solve #x^2+5x+6=0# by factoring?

3 Answers
Jul 16, 2018

Answer:

#x=-3" or "x=-2#

Explanation:

#"the factors of "+6" which sum to "+5" #
#"are "+3" and "+2#

#(x+3)(x+2)=0#

#"equate each factor to zero and solve for x"#

#x+3=0rArrx=-3#

#x+2=0rArrx=-2#

Jul 16, 2018

Answer:

#x=-2# or #x=-3#

Explanation:

Completing the square

#x^2+2*5/2x+25/4+6-25/4=0#

we get

#(x+5/2)^2-1/4=0#

using that #a^2-b^2=(a+b)(a-b)#

we get
#(x+5/2-1/2)(x+5/2+1/2)=0#

so #(x+2)(x+3)=0#

we get #x=-2# or #x=-3#

Jul 16, 2018

Answer:

#x=-2# and #x=-3#

Explanation:

Let's do a little thought experiment:

What two numbers sum to the middle term, and have a product of the last term?

After some trial and error, we arrive at #2# and #3#. This means we can factor this as

#(x+2)(x+3)=0#

Next, we can set both factors equal to zero to get

#x=-2# and #x=-3#

Hope this helps!