Factor using a quadratic pattern

Question

x^4 + 5x^2 -36

Answer= (x+2)(x-2)(x^2+9)

I answered= (x+2)(x-2)(x+3)(x-3)

The quadratic x^2 +9 has no real roots. By the Factor Theorem, it does not have any linear factors with real coefficients.



Could someone please explain this last statement? I feel there is some fundamental principals I am not completely understanding. Also, please feel free to suggest any areas I would need to explore to better understand this type of problem.

Thanks

Answer 1

x^2-9 can be factored into (x+3)(x-3) with real roots 3, -3

but x^2+9 could be viewed as x^ -(-3i)^2, then it factors into (x-3i)(x+3i) with two imaginary roots, 3i and -3i

3i and -3i are not real numbers. Although it isn't usually done that way. Usually, you just take x^2+9 and set it equal to zero, solve for

x^2+9=0

x^2=-9 then take the square root of both sides

x= + or - square root of -9

x= 3i, or -3i

or you could have applied the quadratic formula to solve for x

x= [-b+ or - sqr(b^2-4ac)]/2a b=0, a=1, c=9

x=sqr(-36)/2 = 6/2 sqr(-1) = 3i

the difference of two squares can be factored into the two factors, one is the sum of the numbers, the other is the difference: (a^2-b^2) = (a+b)(a-b)

Possibly you might review a section of your textbook on imaginary numbers?

x^2+9 cannot be factored without resorting to imaginary factors (x-3i)(x+3i)