How do u factorise: I think differnece of 2 squares
9x^2 - 25y^2
36x^2 - 4
2x^2 - 50
2x^4 + 14x^2 + 24
+129839 Note that the first "rule" of factoring is to take out any common factors, first. Once we have done this, if we have a difference of two squares remaining, we just write the square root of both factors twice........separate the first pair with a "+" and the second pair with a "-"
So, the frist one has no common factors.......take the square root of both terms, wtite them twice, and separate them with a "+" and a "-".....so we have
(3x + 5y ) ( 3x - 5y ) ...... and that's the factorization
For the second one, take out the common factor of 4 and we have...
4(9x2 - 1).......now, just factor the inside expression like the first one....don't forget to append the "4" to it when you're finished !!!
The third one is similar to the second one
For the last one, take out the common factor of "2" and we're left with
2(x^4 + 7x^2 + 12)......this isn't a perfect square trinomial, but can be factored as
2(x^2 + 4) (x^2 + 3)
Hope that helps
+129839 Note that the first "rule" of factoring is to take out any common factors, first. Once we have done this, if we have a difference of two squares remaining, we just write the square root of both factors twice........separate the first pair with a "+" and the second pair with a "-"
So, the frist one has no common factors.......take the square root of both terms, wtite them twice, and separate them with a "+" and a "-".....so we have
(3x + 5y ) ( 3x - 5y ) ...... and that's the factorization
For the second one, take out the common factor of 4 and we have...
4(9x2 - 1).......now, just factor the inside expression like the first one....don't forget to append the "4" to it when you're finished !!!
The third one is similar to the second one
For the last one, take out the common factor of "2" and we're left with
2(x^4 + 7x^2 + 12)......this isn't a perfect square trinomial, but can be factored as
2(x^2 + 4) (x^2 + 3)
Hope that helps
