Question

Find all the zeros of the equation.

-4x^4 - 44x^2 + 3600 = 0

A. 5, -5, 6i, -6i

B. 5, 6i

C. 5, -5, 6i, 0

D. -5, -6i

Answer 1

The highest exponent of the variable in this polynomial is 4, so we must have 4 zeros.

In a polynomial with only real numbers as coefficients, any complex roots must come in conjugate pairs \(a \pm bi\) where \(i = \sqrt{-1}\)

A bit of careful consideration will give you the answer from that information alone. However, in a more practical setting you wouldn't have the luxury of choosing from a list of possible answers. In the next post we'll find out how to find them.

Answer 2

If I'm understanding you correctly, I would choose A.

Answer 3

We can factor this polynomial:

\(-4x^4-44x^2+3600 = 0\) by noticing that -4 is a common factor to all of the terms:
\[-4(x^4+11x^2-900) = 0\]Discard the -4 as it doesn't affect the roots:
\[x^4+11x^2-900=0\]
Now consider a substitution: \(u = x^2\)
We can rewrite our polynomial as\[(x^2)*(x^2)+11(x^2) - 900 = 0\]\[u^2+11u-900=0\]Now our task is to factor that quadratic (or solve via completing square or quadratic equation).

Answer 4

If we try to factor it, we want a pair of factors of -900 that add to 11. -25 and 36 is such a pair.
\[u^2+11u-900 = 0\]\[(u-25)(u+36) = 0\]
\(u - 25 = 0\) gives us one pair of solutions
\(u+36 = 0\) gives us the other pair
in both cases, we undo our substitution:
\[x^2-25 = 0\]\[x^2+36=0\]Both of those are differences of squares, so we can solve them by inspection:
\[(x-5)(x+5) = 0\]\[x = \pm 5\]
\[x^2+36 = 0\]\[x^2-i^236 = 0\]\[(x+6i)(x-6i) = 0\]\[x=\pm6i\]

Answer 5

Remember, a difference of squares \[a^2-b^2\]always factors as \((a+b)(a-b)\) because\[(a-b)(a+b) = a^2 + ab - ab - b^2 = a^2-b^2\]If you didn't recognize \(x^2+36\) as being a difference of squares, you could also solve it this way:\[x^2+36=0\]\[x^2=-36\]\[x=\pm\sqrt{-36}\]\[x=\pm\sqrt{-1*36}=\pm\sqrt{-1}*\sqrt{36} = \pm6\sqrt{-1}=\pm6i\]

Answer 6

In general, if you see a polynomial where the successive exponents are common multiples of 2 and 1, such as
\[ax^4 + bx^2+c\]\[ax^6+bx^3+c\]\[ax^8+bx^4+c\]etc. this technique will work.

Answer 7

That was the best explanation I've ever read. Thank you very much.

Answer 8

Thanks! It occurred to me when I finished that I might not exactly have followed the directions in your username, but I'm glad it was clear enough regardless :-)