(x2 + 1)(x - 3)
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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)
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
3x2 - 2x + 3
It is a polynomial if the square root is in a coefficient but not if it is applied to the variable. A polynomial can have only integer powers of the variable. Thus: sqrt(2)*x3 + 4*x + 3 is a polynomial expression but 2*x3 + 4*sqrt(x) + 3 is not.
f(x)=x3-3x2-5x+39=(x+3)(x2-6x+13) It has three roots. One of which is x=-3. Using the quadratic equation: x = (6 +/- √(-16))/2 x = (6 +/- 4i)/2 = (3 +/- 2i) so, x=-3, x=3+2i, or x=3-2i