Yes.
(2x)2 = 2*2*x*x
or
2x2 = 2 * x * x
2x^2-77x+119 It is not factorable if you are looking for integral factors (whole numbers). You might be able to use the quadratic formula if you wish to solve for the zeros (x-intercepts) of it.
4(h2 - 14)
2x squared minus 4
Yes you have to factor it. (3x^2+5)(2x^2-3)
If you mean: (2x-7y)(x+y) then it is 2x squared -5xy -7y squared
2x^2-77x+119 It is not factorable if you are looking for integral factors (whole numbers). You might be able to use the quadratic formula if you wish to solve for the zeros (x-intercepts) of it.
4(h2 - 14)
2x -2 squared = 0
2x squared minus 4
The GCF is 2x.
-x2+2x
When you subtract 7x^2 from 5x^2, you get -2x^2.
Yes you have to factor it. (3x^2+5)(2x^2-3)
2
2x2 + 2x = 2x(x+1)
15
7x2 -4x -5x2 +2x = 2x2 -2x