how do you factor "x^2 + x +xy +y" ?

Question

the book says it's (x + 1)(x + y), but I can't figure out how to get that!

Answer 1

x^2 + x +xy +y ;
Its our function
[ x^2 + x ]+ [ xy +y ]
{ for first part facrot'x ' & for second part factor 'y' }
[ x( x +1) ] + [ y ( x +1) ]
x( x +1) + y ( x +1)
{ here you should factor ' ( x +1) ' }
so we have ;
( x +1)( x + y )
well you see your book is right, and i hope you can figure out now how.

Good Luck.

Answer 2

Group Factoring

x^2 + x + xy + y
(x^2 + x) + (xy + y)
x(x + 1) + y(x + 1)

because both the x and y have the same factor, the problem can factor like this

(x + y)(x + 1) or in your books case (x + 1)(x + y)

Answer 3

you have to factor by grouping
x^2+x+xy+y
(x^2+x)+(xy+y)
pullout an x on from the first set of parentheses, and a y from the second
x(x+1)+y(x+1)
take the common set, then group the x and y that you pulled out
(x+1)(x+y)

hope i explained it well enough
if not here's a site that might help you
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm

PS go down to example 2

Answer 4

x^2 + x + xy + y

==> (x^2 + x) + (xy + y)

believe it or not that is the cricial step.

==> x(x + 1) + y(x + 1)

Do you see? Do you recognize this as something of the form:

xA + yA ??

Then you can write A(x + y)

So you end up with (x + 1)(x + y)

That's how it's done. Subtle, yet obvious extension of the 'example' I gave you above.

I sincerely hope this helps you understand how this works.

Answer 5

the x^2 tells you that you have an x in each bracket.

The xy means that you have a y in one bracket only.

the y on its own and the lack of a coefficient other than 1 tells you that there is a one in the other bracket from the bracket with the y.

This gives (x+1)(x+y) as its factors.

Answer 6

if you use the distributive properto to multiply it then u get the top thing. so like...

first u distribute the first part.
so x*x=x^2

then x*y=xy

then u do everything with the 1.
1*x=x
and
1*y=y

so then u put all those in an expression so its...

x^2+xy+x+y

which is the same thing as "x^2+x+xy+y"

Answer 7

grouping
x^2+x +xy+y
taking out common factors
x(x+1)+y(x+1)
(x+1)(x+y)

Answer 8

x^2+x+xy+y
Factor x out of the first two terms, y out of the second two terms
x(x+1)+y(x+1)
Distributive law in reverse
(x+y)(x+1)