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

3 Answers
Jim G.
Jul 16, 2018

#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#

Sonnhard
Jul 16, 2018

#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#

Jacobi J.
Jul 16, 2018

#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!

Related questions
See all questions in Comparing Methods for Solving Quadratics
Impact of this question
5474 views around the world
You can reuse this answer
Creative Commons License