Question 214080
Factor completely:


{{{2x^2+20x+50=(2x+10)(x+5)}}}


Step 1.  We need to find two integers A and B that A*B=ac and A+B=b where 


{{{ax^2+bx+c}}} in order to be factorable and a=2, b=20 and c=50


Step 2.  In the above example, then A*B=(2)(50)=100 and A+B=20.  


Step 3.  After trial and error the numbers are A=10 and B=10


Step 4.  Using step 3, we can use grouping as follows 


{{{2x^2+20x+50=2x^2+10x+10x+50}}}  where {{{20x=10x+10x}}}


{{{2x^2+20x+50=(2x^2+10x)+(10x+50)}}}


Step 5.  Factor out 2x in first group and 10 in second group


{{{2x^2+10x+50=2x(x+5)+(10)(x+5)}}}


Step 6.  Factor out (x+5) since it is common in both groups.


{{{2x^2+10x+50=(2x+10)(x+5)}}}  ANSWER


We can use the FOIL method to verify


{{{(2x+10)(x+5)=2x^2+10x+10x+50=2x^2+20x+50}}}


I hope the above steps were helpful. 


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J