How do you factor completely $b^{3} + 49 b$ $b^3+49b$ ?

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Answer 1 · 2016-04-21T17:01:26

Real number solution: $b (b^{2} + 49)$ $b(b^2+49)$
Complex number solution: $b (b + 7 i) (b - 7 i)$ $b(b+7i)(b-7i)$

The answer to this question will depend on whether we are allowed to consider imaginary numbers!

Real numbers

We can factor out the common factor of $b$ $b$ which leaves us with:

$b (b^{2} + 49)$ $b(b^2+49)$

There is no way to factor this further which we can check using the discriminant of the quadratic equation $x^{2} + 0 x + 49$ $x^2 + 0x + 49$ :

$Delta = b^2 - 4ac = 0 - 4*49 < 0$

Since it is less than zero, there are no real factors.

Imaginary (complex) numbers

The roots of the quadratic $x^2 + 0x + 49$ are

$x = (-0 +-sqrt(0^2 - 4*49))/(2) = +-7i$

So our expression can be factored into

$b(b+7i)(b-7i)$

How do you factor completely b^3+49bb3+49bb^3+49b?