(x + 1) / (2x3 + 3x + 5) --- Can't depict long division here, but you can work this out by seeing if (x + 1) is a factor of (2x3 + 3x + 5). If it is, then the answer will be 1/(the other factor). In this case, that is true. (2x3 + 3x + 5)/(x + 1) = 2x2 - 2x + 5, so the term term above is equal to: 1 / (2x2 - 2x + 5)
2x2 + 7x + 6 = 2x2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)
You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.You can easily factor this:2x2 + 3x = 0x(2x + 3) = 0Separate this into two equations, one for each solution:x = 02x + 3 = 02x = -3x = -3/2So the two solutions are 0, and -3/2.
2x3 + 8x2 + 3x + 12 = (2x3 + 8x2) + (3x + 12) = 2x2(x + 4) + 3(x + 4) = (x + 4)(2x2 + 3) Since you have asked this question I am assuming that you are not yet at a level where you might want (2x2 + 3) factorised into its imaginary factors.
4x3+12x2+3x+9
(2x + 7)(x - 2)
No. You'd have to use the ABC-formula
2x2-4x-3x+6 2x2-7x+6 (2x+3)(x+2)
(2x + 7)(x - 2)
(3x - 1)(2x2 - 2x - 1)
(2x + 7)(x - 2)
2x2 + 3x + 1 = (x + 1)(2x + 1)
d/dx 2x2+3x+7=4x+3
x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)
2x2 equals 5
2x2-3x-5 = (2x-5)(x+1) when factored
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