How do you factorise #8x^2-6xy-9y^2+10x+21y-12# ?

5 Answers
Sep 18, 2016

Answer:

#E=(2 x - 3 y+4) (4 x + 3 y-3)#

Explanation:

#E# must be of the form #(a_1 x+b_1 y + c_1)(a_2 x+b_2 y + c_2)#

with

#{(a_1a_2=8), (c_1c_2=-12), (b_2 c_1 + b_1 c_2=21), (b_1 b_2=-9), (a_2 c_1 + a_1 c_2=10), (a_2 b_1 + a_1 b_2=-6):}#

Solving for #a_1,b_1,b_2,c_1,c_2# we have as solutions

#a_1 = 8/a_2, b_1 = 6/a_2, b_2= -(3 a_2)/2, c_1 = -6/a_2, c_2 = 2 a_2#

and

#a_1 = 8/a_2, b_1 = -12/a_2, b_2 = (3 a_2)/4, c_1 = 16/a_2, c2 = -(3 a_2)/4#

Choosing #a_2=4# we have

#E=(2 x - 3 y+4) (4 x + 3 y-3)#

or

#E = (2 x + (3 y)/2-3/2)(4 x - 6 y+8)#

Sep 18, 2016

Answer:

#8x^2-6xy-9y^2+10x+21y-12= (4x+3y-3)(2x-3y+4)#

Explanation:

Here's another way of approaching the problem...

Given:

#8x^2-6xy-9y^2+10x+21y-12#

Note that this is a mixture of terms of degree #2#, #1# and #0#, so the factorisation we are looking for will take the form:

#8x^2-6xy-9y^2+10x+21y-12=(ax+by+c)(dx+ey+f)#

By considering the way that terms of different degree multiply, we must also have the simpler:

#8x^2-6xy-9y^2 = (ax+by)(dx+ey)#

Use an AC method to find this factorisation:

Look for a pair of factors of #AC = 8*9 = 72# with difference #B=6#

The pair #12, 6# works.

Use this pair to split the middle term and factor by grouping:

#8x^2-6xy-9y^2 = (8x^2-12xy)+(6xy-9y^2)#

#color(white)(8x^2-6xy-9y^2) = 4x(2x-3y)+3y(2x-3y)#

#color(white)(8x^2-6xy-9y^2) = (4x+3y)(2x-3y)#

Going back to our original problem, we are looking for #c# and #f# such that:

#8x^2-6xy-9y^2+10x+21y-12#

#= (4x+3y+c)(2x-3y+f)#

#= 8x^2-6xy-9y^2 + c(2x-3y) + f(4x+3y) + cf#

#= 8x^2-6xy-9y^2 + 2(c+2f)x+3(f-c)y + cf#

Equating coefficients, we find:

#{ (c + 2f = 5), (f-c=7), (cf=-12) :}#

Adding the first two of these equation, we find:

#3f = 12#

and hence:

#f = 4#

Then:

#c = -12/f = -12/4 = -3#

So:

#8x^2-6xy-9y^2+10x+21y-12= (4x+3y-3)(2x-3y+4)#

Sep 18, 2016

Answer:

#E=(4x+3y-3)(2x-3y+4)#.

Explanation:

We assume that #E# can be factorised into two linear polys. in

#x, &, y#. However, we can verify, using a Test which I will give at

the end, that #E# is factorisable.

Method I :-

We have, #E=8x^2-6xy-9y^2+10x+21y-12#

#=ul(8x^2-12xy)+ul(6xy-9y^2)+10x+21y-12#

#=4x(2x-3y)+3y(2x-3y)+10x+21y-12#

#=(4x+3y)(2x-3y)+10x+21y-12#.

Let #4x+3y=a, and, 2x-3y=b#.

Solving for #x & y, x=(a+b)/6, &, y=(a-2b)/9#.

#:. E=ab+10(a+b)/6+21(a-2b)/9-12#

#=ab+5/3(a+b)+7/3(a-2b)-12#

#=1/3{3ab+5(a+b)+7(a-2b)-36}#

#=1/3{ul(3ab+12a)-ul(9b-36)}#

#=1/3{3a(b+4)-9(b+4)}#

#=1/3(3a-9)(b+4)#

#=(a-3)(b+4)#

#rArr E=(4x+3y-3)(2x-3y+4)#.

Method II :-

In this Method, we proceed, more or less as in Method I, but, instead

of solving for #x,y#, we try to find #l,m in RR#, such that,

#10x+21y=l(4x+3y)+m(2x-3y) rArr l=4, m=-3#.

Hence, #E=ul(ab+4a)-ul(3b-12)=a(b+4)-3(b+4)#

#=(a-3)(b-4)=(4x+3y-3)(2x-3y+4),# as Before!

Enjoy maths.!

Note :-

Method III and the Test will be described in a separate Answer

Sep 18, 2016

Answer:

#(4x+3y-3)(2x-3y+4)#.

Explanation:

Method III :-

#E=8x^2-6xy-9y^2+10x+21y-12#

#=-9y^2-6y(x-7/2)+8x^2+10x-12#

Completing the Square on the basis of the first two terms, we get,

#E=ul{-9y^2-6y(x-7/2)-(x-7/2)^2}+(x-7/2)^2+8x^2+10x-12#

#=-{3y+(x-7/2)}^2+(x^2-7x+49/4)+8x^2+10x-12#

#=-{3y+(x-7/2)}^2+9x^2+3x+1/4#

#=-{3y+(x-7/2)}^2+(3x+1/2)^2#

#=[(3x+1/2)+{3y+(x-7/2)}][3x+1/2-{3y+(x-7/2)}]#

#=(4x+3y-3)(2x-3y+4)#, as Before!

The Test :-

#"The General Second Degree Eqn. in "x & y :#

#S : ax^2+2hxy+by^2+2gx+2fy+c# can be factorised into two

linear polys. iff # |(a,h,g),(h,b,f),(g,f,c)|=0#.

In our case, # |(a,h,g),(h,b,f),(g,f,c)|#

#=|(8,-3,5),(-3,-9,21/2),(5,21/2,-12)|#

#=8(108-441/4)+3(36-105/2)+5(-63/2+45)#

#=8(-9/4)+3(-33/2)+5(27/2)#

#=1/2(-36-99+135)#

#=0#.

Hence, the given poly. #E=E(x,y)# is factorisable.

Enjoy Maths.!

Sep 18, 2016

Answer:

#E=(4x+3y-3)(2x-3y+4)#.

Explanation:

This Final Method is so simple and practical that I can't resist myself

from presenting it.

In this Method, we factorise #3" trinomials from "E# and make a

guess of the factors :

From #E#, we select, #8x^2-6xy-9y^2#, and factorise it as,

#8x^2-6xy-9y^2=(2x-3y)(4x+3y)...............................(1)#

Next,

#8x^2+10x-12=(x+2)(8x-6)=(2x+4)(4x-3).......(2)#

And,

#-9y^2+21y-12=(y-1)(-9y+12)=(3y-3)(-3y+4)..........................................(3)#

Now, comparing #(1),(2),&(3)#, we can guess that,

#E=(4x+3y-3)(2x-3y+4)#.

Of course, we have to verify that the factorisation is right.

Enjoy Maths.!