How do you factor #y=(5x+2)^2 + 11(5x+2)+30 #?

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Answer 1 · 2016-01-18T10:45:44

#color(green)(y=(5x+7)(5x+8))#

Have a look at the 'trick' I used!

Multiply out the brackets so that you can collect like terms and simplify:

Let #(5x+2)# be z then we have

#y=z^2+11z+30#

Factors of 30 are: {1,30} ; {2, 15} ; {3,10} ; {5,6}

We observe that 5+6=11 so we try those first:

#y=(z+5)(z+6) =z^2+6z+5z+30 color(red)(" This works")#

Substitute back for z giving:

#y=[(5x+2)+5)([5x+2]+6)#

#color(green)(y=(5x+7)(5x+8))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check by multiplying out the brackets

#y=(5x+7)(5x+8)#

#y= 25x^2+40x+35x+56color(white)(...) = 25x^2+75x+56 color(red)(" This works")#