# How do you factor completely b^3+49b?

Apr 21, 2016

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

#### Explanation:

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 $b$ which leaves us with:

$b \left({b}^{2} + 49\right)$

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

$\Delta = {b}^{2} - 4 a c = 0 - 4 \cdot 49 < 0$

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

Imaginary (complex) numbers

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

$x = \frac{- 0 \pm \sqrt{{0}^{2} - 4 \cdot 49}}{2} = \pm 7 i$

So our expression can be factored into

$b \left(b + 7 i\right) \left(b - 7 i\right)$