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(3x - 1)(x + 1) or (3x + 1)(x - 1) assuming the 1 is negative.

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Q: Can you solve 3x2 2x 1 by factor?
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What is the factor of 3x2 2x 2 equals 0?

3x2 + 2x + 2 = 0 Can not be factored. You can however solve it for x: 3x2 + 2x = -2 x2 + 2x/3 = -2/3 x2 + 2x/3 + 1/9 = -2/3 + 1/9 (x + 1/9)2 = -5/9 x + 1/9 = ±√(-5/9) x = -1/9 ± i√(5/9) x = -1/9 ± i√5 / 3 x = -1/9 ± 3i√5 / 9 x = (-1 ± 3i√5) / 9


Factor 3x to the third power minus 3x to the second power plus 2x minus 2?

3x3 - 3x2 + 2x - 2 = 3x2(x - 1) + 2(x - 1) = (3x2 + 2)(x - 1)


How do you solve x2 plus 3x4?

This is an expression and you do not solve an expression, but you can factor this one.X2 + 3X4X2(1 + 3X2)========


What is the quotient 3x3 - x2 - x - 1 x - 1?

3x3 - x2 - x - 1 = 3x3 - 3x2 + 2x2 - 2x + x - 1 = 3x2(x - 1) + 2x(x - 1) + 1(x - 1) = (3x2 + 2x + 1)(x - 1) So 3x3 - x2 - x - 1 /(x - 1) = (3x2 + 2x + 1)


What is the name of this function 4y plus 8 equal to 3x to the power of 2 minus 2x plus 1?

if you rearrange it, it becomes 4y + 8 = 3x2 -2x +1 4y = 3x2 -2x -7 y= (3x2 -2x -7)/4 which is a parabola


How do you factor 6x - 6x -16?

6x-6x=0 so all you have left is -16 and there is nothing to factor. I am guessing you meant 6x2 -6x-16 which is 2(3x2 -2x-8) The factor pairs for 8 are 1,8 and 2,4. So let's try 1, 8. We know one of them has to be negative. We see it does not work and when we try 2 and 4 they do not work also. So we cannot factor (3x2 -2x-8) over the integers and we call it prime. The final answer is 2(3x2 -2x-8)


How do you factor 12x2-3x2?

While it is possible to factor 3x2 from both of these and get 3x2(4 - 1), it's a lot easier to subtract 3x2 from 12x2 and get 9x2


What is the factored form of 3x2-2x-1?

It is: (3x+1)(x-1)


What is the answer when 12x3 14x2 14x 2 is divided by 4x plus 2.help please you keep getting a decimal answer?

(12x3 + 14x2 + 14x + 2)/(4x + 2) simplify= (6x3 + 7x2 + 7x + 1)/(2x + 1)1. 6x3 + 7x2 - 3x2(2x + 1) = 6x3 - 6x3 + 7x2 - 3x2 = 4x2; bring down 7x2. 4x2 + 7x - 2x(2x + 1) = 4x2 - 4x2 + 7x - 2x = 5x bring down 13. 5x + 1 - 2(2x + 1) = 5x - 4x + 1 - 2 = x - 1 the reminderThus, (6x3 + 7x2 + 7x + 1)/(2x + 1) = 3x2 + 2x + 2 + (x - 1)/(2x + 1)and(12x3 + 14x2 + 14x + 2)/(4x + 2) = 3x2 + 2x + 2 + (2x - 2)/(4x + 2).


When x3 plus 3x2-2x plus 7 is divided by x plus 1 what is the remainder?

(x3 + 3x2 - 2x + 7)/(x + 1) = x2 + 2x - 4 + 11/(x + 1)(multiply x + 1 by x2, and subtract the product from the dividend)1. x2(x + 1) = x3 + x22. (x3 + 3x2 - 2x + 7) - (x3 + x2) = x3 + 3x2 - 2x + 7 - x3 - x2 = 2x2 - 2x + 7(multiply x + 1 by 2x, and subtract the product from 2x2 - 2x + 7)1. 2x(x + 1) = 2x2 + 2x2. (2x2 - 2x + 7) - (2x2 + 2x) = 2x2 - 2x + 7 - 2x2 - 2x = -4x + 7(multiply x + 1 by -4, and subtract the product from -4x + 7)1. -4(x + 1) = -4x - 42. -4x + 7 - (-4x - 4) = -4x + 7 + 4x + 4 = 11(remainder)


Can you multiply 3 consecutive numbers and get 2015?

This is the same as asking is there an integer solution to x(x+1)(x+2)=2015.x3 -3x2 +2x=2015So we must solve x3 -3x2 +2x-2015=0 for integer solutions.The solution is x=13.657 which is not an integer. That is the only real solution.So the answer is no.


Find the sum 3x2 plus 2x plus 3 plus x2 plus x plus 1?

3x2 + 2x + 3 + x2 + x + 1 = 4x 2+ 3x + 4