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X^2-9x-4

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Anonymous

5y ago

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What is 2x squared plus x2 equal?

2x2 + x2 = 3x2


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)


What are similar terms?

They are terms that have the same variables like x, 2x and 3x or x2, 2x2 and 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 must be subtracted from 4x power 4 so that the result is exactly divisible by 2x power 2 plus x-1?

NOT CALCULUS. Use long division. 2x2 + x - 1(4x^4) 2x^2 goes into 4x^4 2x^2 times. The remainder will then be (2x^2*(x-1)). This result(2x^3 - 2x^2) is what need to be subtracted from 4x^4 to make it exactly divisible


What are similar algebraic terms?

They are terms that have the same variables like x, 2x and 3x or x2, 2x2 and 3x2


What is 2x squared plus 5x-4x squared plus x-x squared?

2x2 + (5x-4x)2 + (x-x)2 = 2x2 + x2 + 02 = 2x2 + x2 = 3x2


What is x squared minus x squared?

x2 - (2x)2 = x2 - 4x2 = -3x2 x2 - 2x2 = -x2


How do you factor out 2x2 -2x?

2x2- 2x = 2x*(x - 1)


What is 2x to the power of 2 minus x to the power of 2?

The question is ambiguous and two possible answers are given below: 2x2 - x2 = x2 or (2x)2 - x2 = 4x2 - x2 = 3x2


What is the result when 8x 3 is subtracted from negative 2x 5?

It is negative 34.


What is the sum of 5x4 - 3x2 - 2x 2x4 2x3 2x2 x 1?

7x^4+2x^3-x^2-x+1 (Apex ;D)If for some reason this isn't the answer then I am very sorry...