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What is the solution to x3 plus 125 equals 0?
It is x = -5
How do you factor 125 -x cubed?
Difference of two cubes: a3 - b3 = (a-b)(a2+ab+b2) 125 - x3 = 53 - x3 = (5 - x)(52 + 5x + x2) = (5 - x)(25 + 5x + x2)
How do you solve x3 plus 125 is equals to 0?
x^(3) + 125 = 0 Remember that 125 = 5^(3) Hence x^(3) + 5^(3) = 0 Factor (x + 5) ( x^(2) - 5x + 25) = 0 So x = -5 & x^(2) - 5x + 125 = 0 Apply Quadratic Eq'n x = {--5 +/- sqrt[)=5)^(2) - 4(1)(125)]} / 2(1) x = { 1 +/- sqrt[25 - 500]} / 2 x = { 1 +/- sqrt[-374] } / 2 This part remain unresolved because you cannot rake the square root of a negative number. Hence x = -5 is the only result.
Can x cubed plus 125 be simlified?
x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)
What is x3 plus 125?
It is a cubic polynomial in x and its value depends on the value of x.
What is the solution to x3 plus 125 equals 0?
It is x = -5
How do you factor 125 -x cubed?
Difference of two cubes: a3 - b3 = (a-b)(a2+ab+b2) 125 - x3 = 53 - x3 = (5 - x)(52 + 5x + x2) = (5 - x)(25 + 5x + x2)
How do you solve x3 plus 125 is equals to 0?
x^(3) + 125 = 0 Remember that 125 = 5^(3) Hence x^(3) + 5^(3) = 0 Factor (x + 5) ( x^(2) - 5x + 25) = 0 So x = -5 & x^(2) - 5x + 125 = 0 Apply Quadratic Eq'n x = {--5 +/- sqrt[)=5)^(2) - 4(1)(125)]} / 2(1) x = { 1 +/- sqrt[25 - 500]} / 2 x = { 1 +/- sqrt[-374] } / 2 This part remain unresolved because you cannot rake the square root of a negative number. Hence x = -5 is the only result.
Can x cubed plus 125 be simlified?
x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)
What x5 equals 125?
To find what x5 equals 125, you can set up the equation 5x = 125. To solve for x, divide both sides by 5: x = 125 / 5 = 25. Therefore, x5 equals 125 when x is equal to 25.
How do you factor x to the third power plus 125?
x^(3) + 125 => x^(3) + 5^(3) This factors to (x + 5)(x^(2) - 5x + 5^(2) ) or (x + 5)(x^(2) - 5x + 25) NB The clue is to spot that the constant (125) is a cubic numbers ( 5^(3)). Similarly with other 'cubic factorings'.
What is x3 plus 125?
It is a cubic polynomial in x and its value depends on the value of x.
Prime factorization for 125?
We note that '125' ends in '5'. So it will divide by '5', which is a prime number. Hence 5)125 5)25 5)5 = 1 Keep dividing by any prime number, that will go into the answer, in this case '5'. So 5 x 5 x 5 = 5^(3) = 125 NB Any number that ends in '5' or '0' will divide by '5'. Any number that ends with an EVEN number will divide by '2'. Any number were the digits sum to '9' will divide by '9'. e.g. 27 ; 2 + 7 = 9 27/9 = 3
How do you divide an exponential equation?
X5/X3 5 - 3 = X2 =====That is the exponent part; subtract the bottom exponent from the top exponent. X3/X5 3 - 5 X - 2 =====or, 1/X2 ===
Factor the polynomial of x3 plus 5x2 - x - 5?
x3 + 5x2 - x - 5 = (x2 - 1)(x + 5) = (x + 1)(x - 1)(x + 5)
What is 125 of 5?
5 is a factor of 125 5 x 5 x 5 = 125
Simplify 9x 5 - x 3?
I am trying to read this right: (9 x 5) - x3 = ? or 9 x (5 - x3) = ? Which one is it?
