How do you factor # 9x^2+24xy+16y^2#?

2 Answers
Mar 17, 2018

Answer:

#9x^2+24xy+16y^2 = (3x+4y)^2#

Explanation:

Note that:

#9x^2 = (3x)^2#

#24xy = 2(3x)(4y)#

#16y^2 = (4y)^2#

So:

#9x^2+24xy+16y^2 = (3x)^2+2(3x)(4y)+(4y)^2#

is in the form:

#A^2+2AB+B^2 = (A+B)^2#

So putting #A=3x# and #B=4y# we have:

#9x^2+24xy+16y^2 = (3x+4y)^2#

Mar 17, 2018

Answer:

#(3x+4y)(3x+4y)#

Explanation:

We are given:

#9x^2 + 24xy + 16y^2#

We want to obtain an expression of the form:

#(ax+by)(cx+dy)#

where #a,b,c,d# are integers (not necessarily unique from each other).

Expanding this form we get:

#acx^2 + adxy +bcxy + bdy^2#

#acx^2 +(ab+cd)xy + bdy^2#

From the expression we are given we must satisfy the following equations:

#ac = 9#
#bd = 16#
#ab+cd = 24#

For #(a,c)# we can have (eliminating any repeats):
#(1,9),(3,3),cancel{(9,1)}#

For #(b,d)# we can have (eliminating any repeats):
#(1,16),(2,8),(4,4),cancel{(8,2)},cancel{(16,1)}#

With these options, we now need to find the ones that when combined will give us the #ab+cd = 24#:

#(1,9)(1,16) -> 1xx1+9xx16=145#
#(1,9)(2,8) -> 1xx2+9xx8 =74#
#(1,9)(4,4) -> 1xx4+9xx4 = 40#
#(3,3)(1,16) -> 3xx1+3xx16= 51#
#(3,3)(2,8) -> 3xx2+3xx8=30#
#color(blue)((3,3)(4,4) -> 3xx4+3xx4 =24)#

Hence, we must choose #(a,c) = (3,3)# and #(b,d) = (4,4)#

So the factored expression #(ax+by)(cx+dy)# is:

#9x^2 + 24xy + 16y^2= color(green){(3x+4y)(3x+4y)}#