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

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Answer 1 · 2016-04-21T17:01:26

Real number solution: #b(b^2+49)#
Complex number solution: #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# which leaves us with:

#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 + 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)#