How do you factor the trinomial #21x^2-40xy+11y^2#?

1 Answer
Nov 24, 2015

#color(green)((3x-y)(7x-11y))#

Explanation:

The factors for 21 are {3, 7}, {1, 21}
The factors for 11 are {1, 11}

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A quick mental check shows that {1, 21} for 21 will cause problems for the rest so it is rejected. Consequently most of the work is already done for you!

For #21x^2# It has to be either #{-3x xx -7x} "or" {+3x xx +7x}#

I am going to choose:

#(3x +?)(7x+?)#

Now lets look at the #11y^2#

This can only be the positive or negative versions of #{1yxx11y}#
#color(brown)("Lets play!")#
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#color(blue)(Try color(white)(.) 1)#

#(3x-11y)(7x-y) = 21x^2 -3xy -77xy +11y^2 #
#color(red)("This fails for the target of "-40xy)#

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#color(blue)(Try color(white)(.) 2)#

#color(green)((3x-y)(7x-11y)) = 21x^2 -33xy-7xy+11y^2#

#color(blue)(=21x^2-40xy+11y^2) color(white)(...) color(red)( "It works!")#