The answer would be 3x if 'X2' is '2x', but if 'X2' is x2, then the total answer would be x2 + x.
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
x4 +x2 =x2 (x2+1)
If 'x' is more than 1, then x2 is more than 'x'. If 'x' is less than 1, then x2 is less than 'x'.
-167
x2-x=x(x-1)
The answer would be 3x if 'X2' is '2x', but if 'X2' is x2, then the total answer would be x2 + x.
The factors of x2 are x and x x times x plus 1 = x2 + x
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
The factors of x2 are x and x.
X3 X(X2) X2(X) and, X * X * X
Answer: x (x2-x)
x2 + 49 = x2 - (-49) = x2 - (-1)(49) = x2 - (i2)(72) = x2 - (7i)2 = (x - 7i)(x + 7i) where i is the imaginary square root of -1.
x2 - (2x)2 = x2 - 4x2 = -3x2 x2 - 2x2 = -x2
x2(x-4)
-x2 = x - 6 x2 + x - 6 = 0 (x - 2)(x + 3) = 0 x ∈ {-3, 2}
x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.