The idea is to add like terms together - in this case, the terms that have the variable "x" raised to the same power.
To find the original number when given the prime factors, we have to work backwards. First, we simplify into 2 squared times 3 squared. (2x2) x (3x3) Then we expand the brackets. 4 x 9 = 54 Therefore 54 is the number, with 2x2 and 3x3 as the prime factors.
4 (2x2) 9 (3x3) and 16 (4x4)
2x2=4 1x4=4 3x3=9 1x9=9
Yes I can. I did it in QBasic about 15 years ago.
1x1 = 1 2x2 = 4 3x3 = 9 4x4 = 16 ... Up to 31x31
17 2x4=8+(3x3)=17 3x3=9 9+8=17
To find the original number when given the prime factors, we have to work backwards. First, we simplify into 2 squared times 3 squared. (2x2) x (3x3) Then we expand the brackets. 4 x 9 = 54 Therefore 54 is the number, with 2x2 and 3x3 as the prime factors.
Multiply the number by itself: 2x2=4, 3x3=9 and so on.
Multiply them. The number is 36.
No, it isn't. You can express 3x3-2x2 as 3x3-2x2+0x+0, so it actually has four terms. The definition of a binomial is an expression in the form Ax+b, where A and b are constants, so 3x3-2x2 is not a binomial. It is actually a quartomial.
72 ^ 9x8 ^ ^ 3x3 4x2 ^ 2x2
If you mean '2x4' as in 2 times 4- 2x4= 8 3x3= 9 4x2= 8 8-9-8=-19 if you want to divide by x -b/c x doesn't have a set amount- it would be: -19/x
4 (2x2) 9 (3x3) and 16 (4x4)
put 4 refined wood in the 2x2 grid in a square
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
i do not understand your question, do you mean: 2x2=4? 3x3=9? 4x4=16?
7x(3x3 - 2x2 + 5x + 1)