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x4 - 1.

We can not "solve" this as we have not been told the value of x. However, we can simplify this expression:

We have an x and a minus x here which will cancel out. Likewise the x2 and x3 will cancel out with the -x2 and -x3 respectively. This therefore leaves us with just x4 - 1.

Q: How do you solve x-1 plus x2-x plus x3-x2 plus x4-x3?

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x equals 4

3

480

what are the coefficient of this number -8a+2=13

6(X-2)-4X=16before you can solve for x you first have to bring the x terms togetherand to do that you distribute the 6 to remove the parenthesis6x-(6)2-4x=16 turns into6x-12-4x=16because -4x and 6x are terms of the same degree they can be added togetheryoull get2x-12=16now solve for x2x-12=16 get x term alone by addind 122x=28 now divide 2x=14

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x equals 4

2x+x2x = x+xso 2x+x isx+x+x = 3x

3

The answer is 9.You solve it as:2x-9=x2x-x=9x=9

480

what are the coefficient of this number -8a+2=13

IF 3X-1=11, WHAT IS THE VALUE OF X 2 + X?

If you mean: 4-x = 2x-5 then the value of x is 3

-if3X-1=11,what is the value of X^2+X?

6(X-2)-4X=16before you can solve for x you first have to bring the x terms togetherand to do that you distribute the 6 to remove the parenthesis6x-(6)2-4x=16 turns into6x-12-4x=16because -4x and 6x are terms of the same degree they can be added togetheryoull get2x-12=16now solve for x2x-12=16 get x term alone by addind 122x=28 now divide 2x=14

e2xlnx = x2x since ln x is the opposite to e so you just leave the x from the log and the rest of the exponent stays the same.

In a quadratic equation, the vertex (which will be the maximum value of a negative quadratic and the minimum value of a positive quadratic) is in the exact center of any two x values whose corresponding y values are equal. So, you'd start by solving for x, given any y value in the function's range. Then, you'd solve for y where x equals the middle value of the two x's given in the previous. For example:y = x24 = x2x = 2, -2y = (0)2y = 0Which is, indeed, the vertex of y = x2