Verified answer

Let u = x^2. Then the equation becomes a simple quadratic equation. This type of substitution is quite common because it lets you change a fourth degree equation into a quadratic. This technique will be used heavily in trig as you will find out when you take trig. I know - I taught for five years while getting my doctorate in mathematics.

u^2 + 34u + 225 = 0

(u + 25) * (u + 9) = 0

So u = -25 and u = -9

So x^2 = -25 and x^2 = -9

x = 5i twice and 3i twice <------------- Roots

(x - 5i) * (x - 5i) * (x - 3i) * (x - 3i) <------------ Factored

With all apologies, I don't use Wolfram Alpha or Maple - I solve the problems on my own. Students told that answers came from sources learn nothing. It's like saying, "Well you been a passenger in a plane, now go fly it". The way I did it is the way it should be done! One step at a time.

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OK, so this is (x^2+9)(x^2+25)

So, it's zero if x^2+9 == 0 x^2 = -9 or x^2 = -25.

This only has zeros where x = +/- 3i or +/- 5i

... f(x) = x^4 + 34x² + 225

or f(x) = [x²]^2 + 34[x²] + 225

or f(x) = ( [x²] + 25 )( [x²] + 9 )

x² = -25 → .. x = ± 5 i where i = √-1

x² = -9 → .... x = ± 3 i where i = √-1

f(x) = ( x + 5i )( x - 5i )( x + 3i )( x - 3i )

b)

factor(x^4+34*x^2+225) = (x^2 + 25)*(x^2 + 9)

a)

x1 = 5*i

x2 = - 5*i

x3 = 3*i

x4 = - 3*i