5
x3-343
X3 X(X2) X2(X) and, X * X * X
Since x3 is a factor of x5, it is automatically the GCF.
Greatest common factor of x4 and x3 is x3.
The greatest common factor of 1-x3 is 1, dummy.
No.
It is x = -5
x2 + 5x + 25
It is a cubic polynomial in x and its value depends on the value of x.
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)
x3 + 125 Use the sum of two cubes. (x+5)(x2-5x+25)
125 * x3 = (5x)3
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.
It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0It could be x3 + 2x + 0
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7
x3
x3-343