Please faCtor. Thank you for your help.
a^2 + 4a + 21
6 answers
typo I think, not negative 21?
NO it's postive 21
Sorry, all you can do is use complex roots then.
Look for the roots of a^2 + 4 a + 21 = 0
a = [-4 +/- sqrt (16-84)]/2
a = [-4 +/- sqrt (-68) ]/2
a = [-2 +/- 2 i sqrt_17 ] / 2
a = -1 + i sqrt 17 or a = -1 + i sqrt 17
so
(a + 1 + i sqrt 17)(a + 1 - i sqrt 17)
Look for the roots of a^2 + 4 a + 21 = 0
a = [-4 +/- sqrt (16-84)]/2
a = [-4 +/- sqrt (-68) ]/2
a = [-2 +/- 2 i sqrt_17 ] / 2
a = -1 + i sqrt 17 or a = -1 + i sqrt 17
so
(a + 1 + i sqrt 17)(a + 1 - i sqrt 17)
Would the answer be no souliton?
There is no real number solution.
Thank you for your help.