Question

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

Answer 1

`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)`

Answer 2

`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)`

Answer 3

`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

Answer 4

`(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.!

Answer 5

`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.!