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

Sep 18, 2016

$E = \left(2 x - 3 y + 4\right) \left(4 x + 3 y - 3\right)$

#### Explanation:

$E$ must be of the form $\left({a}_{1} x + {b}_{1} y + {c}_{1}\right) \left({a}_{2} x + {b}_{2} y + {c}_{2}\right)$

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

${a}_{1} = \frac{8}{a} _ 2 , {b}_{1} = \frac{6}{a} _ 2 , {b}_{2} = - \frac{3 {a}_{2}}{2} , {c}_{1} = - \frac{6}{a} _ 2 , {c}_{2} = 2 {a}_{2}$

and

${a}_{1} = \frac{8}{a} _ 2 , {b}_{1} = - \frac{12}{a} _ 2 , {b}_{2} = \frac{3 {a}_{2}}{4} , {c}_{1} = \frac{16}{a} _ 2 , c 2 = - \frac{3 {a}_{2}}{4}$

Choosing ${a}_{2} = 4$ we have

$E = \left(2 x - 3 y + 4\right) \left(4 x + 3 y - 3\right)$

or

$E = \left(2 x + \frac{3 y}{2} - \frac{3}{2}\right) \left(4 x - 6 y + 8\right)$

Sep 18, 2016

$8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12 = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$

#### Explanation:

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

Given:

$8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12$

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:

$8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12 = \left(a x + b y + c\right) \left(\mathrm{dx} + e y + f\right)$

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

$8 {x}^{2} - 6 x y - 9 {y}^{2} = \left(a x + b y\right) \left(\mathrm{dx} + e y\right)$

Use an AC method to find this factorisation:

Look for a pair of factors of $A C = 8 \cdot 9 = 72$ with difference $B = 6$

The pair $12 , 6$ works.

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

$8 {x}^{2} - 6 x y - 9 {y}^{2} = \left(8 {x}^{2} - 12 x y\right) + \left(6 x y - 9 {y}^{2}\right)$

$\textcolor{w h i t e}{8 {x}^{2} - 6 x y - 9 {y}^{2}} = 4 x \left(2 x - 3 y\right) + 3 y \left(2 x - 3 y\right)$

$\textcolor{w h i t e}{8 {x}^{2} - 6 x y - 9 {y}^{2}} = \left(4 x + 3 y\right) \left(2 x - 3 y\right)$

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

$8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12$

$= \left(4 x + 3 y + c\right) \left(2 x - 3 y + f\right)$

$= 8 {x}^{2} - 6 x y - 9 {y}^{2} + c \left(2 x - 3 y\right) + f \left(4 x + 3 y\right) + c f$

$= 8 {x}^{2} - 6 x y - 9 {y}^{2} + 2 \left(c + 2 f\right) x + 3 \left(f - c\right) y + c f$

Equating coefficients, we find:

$\left\{\begin{matrix}c + 2 f = 5 \\ f - c = 7 \\ c f = - 12\end{matrix}\right.$

Adding the first two of these equation, we find:

$3 f = 12$

and hence:

$f = 4$

Then:

$c = - \frac{12}{f} = - \frac{12}{4} = - 3$

So:

$8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12 = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$

Sep 18, 2016

$E = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$.

#### 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, $E = 8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12$

$= \underline{8 {x}^{2} - 12 x y} + \underline{6 x y - 9 {y}^{2}} + 10 x + 21 y - 12$

$= 4 x \left(2 x - 3 y\right) + 3 y \left(2 x - 3 y\right) + 10 x + 21 y - 12$

$= \left(4 x + 3 y\right) \left(2 x - 3 y\right) + 10 x + 21 y - 12$.

Let $4 x + 3 y = a , \mathmr{and} , 2 x - 3 y = b$.

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

$\therefore E = a b + 10 \frac{a + b}{6} + 21 \frac{a - 2 b}{9} - 12$

$= a b + \frac{5}{3} \left(a + b\right) + \frac{7}{3} \left(a - 2 b\right) - 12$

$= \frac{1}{3} \left\{3 a b + 5 \left(a + b\right) + 7 \left(a - 2 b\right) - 36\right\}$

$= \frac{1}{3} \left\{\underline{3 a b + 12 a} - \underline{9 b - 36}\right\}$

$= \frac{1}{3} \left\{3 a \left(b + 4\right) - 9 \left(b + 4\right)\right\}$

$= \frac{1}{3} \left(3 a - 9\right) \left(b + 4\right)$

$= \left(a - 3\right) \left(b + 4\right)$

$\Rightarrow E = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$.

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 $l , m \in \mathbb{R}$, such that,

$10 x + 21 y = l \left(4 x + 3 y\right) + m \left(2 x - 3 y\right) \Rightarrow l = 4 , m = - 3$.

Hence, $E = \underline{a b + 4 a} - \underline{3 b - 12} = a \left(b + 4\right) - 3 \left(b + 4\right)$

$= \left(a - 3\right) \left(b - 4\right) = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right) ,$ as Before!

Enjoy maths.!

Note :-

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

Sep 18, 2016

$\left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$.

#### Explanation:

Method III :-

$E = 8 {x}^{2} - 6 x y - 9 {y}^{2} + 10 x + 21 y - 12$

$= - 9 {y}^{2} - 6 y \left(x - \frac{7}{2}\right) + 8 {x}^{2} + 10 x - 12$

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

$E = \underline{- 9 {y}^{2} - 6 y \left(x - \frac{7}{2}\right) - {\left(x - \frac{7}{2}\right)}^{2}} + {\left(x - \frac{7}{2}\right)}^{2} + 8 {x}^{2} + 10 x - 12$

$= - {\left\{3 y + \left(x - \frac{7}{2}\right)\right\}}^{2} + \left({x}^{2} - 7 x + \frac{49}{4}\right) + 8 {x}^{2} + 10 x - 12$

$= - {\left\{3 y + \left(x - \frac{7}{2}\right)\right\}}^{2} + 9 {x}^{2} + 3 x + \frac{1}{4}$

$= - {\left\{3 y + \left(x - \frac{7}{2}\right)\right\}}^{2} + {\left(3 x + \frac{1}{2}\right)}^{2}$

$= \left[\left(3 x + \frac{1}{2}\right) + \left\{3 y + \left(x - \frac{7}{2}\right)\right\}\right] \left[3 x + \frac{1}{2} - \left\{3 y + \left(x - \frac{7}{2}\right)\right\}\right]$

$= \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$, as Before!

The Test :-

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

$S : a {x}^{2} + 2 h x y + b {y}^{2} + 2 g x + 2 f y + c$ can be factorised into two

linear polys. iff $| \left(a , h , g\right) , \left(h , b , f\right) , \left(g , f , c\right) | = 0$.

In our case, $| \left(a , h , g\right) , \left(h , b , f\right) , \left(g , f , c\right) |$

$= | \left(8 , - 3 , 5\right) , \left(- 3 , - 9 , \frac{21}{2}\right) , \left(5 , \frac{21}{2} , - 12\right) |$

$= 8 \left(108 - \frac{441}{4}\right) + 3 \left(36 - \frac{105}{2}\right) + 5 \left(- \frac{63}{2} + 45\right)$

$= 8 \left(- \frac{9}{4}\right) + 3 \left(- \frac{33}{2}\right) + 5 \left(\frac{27}{2}\right)$

$= \frac{1}{2} \left(- 36 - 99 + 135\right)$

$= 0$.

Hence, the given poly. $E = E \left(x , y\right)$ is factorisable.

Enjoy Maths.!

Sep 18, 2016

$E = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$.

#### Explanation:

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

from presenting it.

In this Method, we factorise $3 \text{ trinomials from } E$ and make a

guess of the factors :

From $E$, we select, $8 {x}^{2} - 6 x y - 9 {y}^{2}$, and factorise it as,

$8 {x}^{2} - 6 x y - 9 {y}^{2} = \left(2 x - 3 y\right) \left(4 x + 3 y\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

Next,

$8 {x}^{2} + 10 x - 12 = \left(x + 2\right) \left(8 x - 6\right) = \left(2 x + 4\right) \left(4 x - 3\right) \ldots \ldots . \left(2\right)$

And,

$- 9 {y}^{2} + 21 y - 12 = \left(y - 1\right) \left(- 9 y + 12\right) = \left(3 y - 3\right) \left(- 3 y + 4\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right)$

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

$E = \left(4 x + 3 y - 3\right) \left(2 x - 3 y + 4\right)$.

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

Enjoy Maths.!