Factorise..
x² + xy + 8x + 8y
With full explanation... Plz plz plz plz
Answers
Answered by
299
Hello...
Here is your answer..
x² + xy + 8x + 8y
We observe that there is no common factor among all terms. Also, there are four terms.
So, let us think of grouping the terms in pairs in such a way that there are some factors common to them and after taking factors common from each pair same binomial is left inside the two brackets. We observe that first two terms have x as common factor. Taking x common from them, we have
x² + xy = x ( x + y )
Also, 8 is a common factor from the Last two terms. Taking 8 common from the last two terms, we have
8x + 8y = ( x + y)
Clearly, x + y is common from the two groups.
Thus, we group the two terms as follows :
x² + xy + 8x + 8y = (x² + xy) + (8x + 8y)
= x ( x + y) + 8 ( x + y)
= (x + 8) ( x + y)
Hope it helped ☺☺
Here is your answer..
x² + xy + 8x + 8y
We observe that there is no common factor among all terms. Also, there are four terms.
So, let us think of grouping the terms in pairs in such a way that there are some factors common to them and after taking factors common from each pair same binomial is left inside the two brackets. We observe that first two terms have x as common factor. Taking x common from them, we have
x² + xy = x ( x + y )
Also, 8 is a common factor from the Last two terms. Taking 8 common from the last two terms, we have
8x + 8y = ( x + y)
Clearly, x + y is common from the two groups.
Thus, we group the two terms as follows :
x² + xy + 8x + 8y = (x² + xy) + (8x + 8y)
= x ( x + y) + 8 ( x + y)
= (x + 8) ( x + y)
Hope it helped ☺☺
Anonymous:
hay dear
Answered by
257
hay!!
=> x²+xy+8x+8y
=> (x²+xy) +(8x+8y)
=> x(x+y)+8(x+y)
=> (x+8) (x+y)
I hope it's help you
=> x²+xy+8x+8y
=> (x²+xy) +(8x+8y)
=> x(x+y)+8(x+y)
=> (x+8) (x+y)
I hope it's help you
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