How do you factorise #8x^2-6xy-9y^2+10x+21y-12# ?
5 Answers
Explanation:
with
Solving for
and
Choosing
or
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
#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
The pair
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
#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)#
Explanation:
We assume that
the end, that
Method I :-
We have,
Let
Solving for
Method II :-
In this Method, we proceed, more or less as in Method I, but, instead
of solving for
Hence,
Enjoy maths.!
Note :-
Method III and the Test will be described in a separate Answer
Explanation:
Method III :-
Completing the Square on the basis of the first two terms, we get,
The Test :-
linear polys. iff
In our case,
Hence, the given poly.
Enjoy Maths.!
Explanation:
This Final Method is so simple and practical that I can't resist myself
from presenting it.
In this Method, we factorise
guess of the factors :
From
Next,
And,
Now, comparing
Of course, we have to verify that the factorisation is right.
Enjoy Maths.!