How do you factor #y= 5x^2 + 22x + 8# ?

1 Answer
Yonas Yohannes
Jan 19, 2016

Ok write the quadratic using factors, you may try
#(5x+2)(x+4)#multiply out and test...
Key is to think of the quadratic as
#(ax+h)(x+k)# multiply long form

#ax^2+hx+akx+hk#
#ax^2+(h+ak)x+hk #

Now see #a = 5; h=2 ; k=4 #
So tha h+ak = 2+20=22; and kh= 8
As required by your quadratic

Hope it was helpful, Yonas

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