Yes.
(2x)2 = 2*2*x*x
or
2x2 = 2 * x * x
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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