Answer:
Step-by-step explanation:
There are 6 solutions or zeros here because, according to the Fundamental Theorem of Algebra, the degree of the polynomial dictates how many zeros there are in the polynomial. If we had a 3rd degree polynomial, we would expect to find 3 zeros; if we had a 5th degree polynomial, we would have 5 zeros, etc. The easiest way to factor this is to do it initially by grouping:
[tex](4x^6+20x^4)-(25x^2-125)[/tex] then
[tex]4x^4(x^2+5)-25(x^2+5)[/tex] then
[tex](x^2+5)(4x^4-25)=0[/tex]
We will factor each set of parenthesis now to get all the zeros. For the first set of parenthesis:
[tex]x^2+5=0[/tex] so
[tex]x^2=-5[/tex] so
[tex]x=\sqrt{-5}[/tex]
But since we can't have a negative under the square root, we have to offset it by using the imaginary number i. i-squared = -1, so
x = ±i√5
Those are the first 2 zeros out of 6. Now for the second set of parenthesis:
4x⁴ - 25 = 0. That is the difference between perfect squares, and that factors to this:
(2x² + 5)(2x² - 5)
The first set of parenthesis there:
2x² + 5 = 0 so
2x² = -5 so
x² = -5/2 so
x = ±[tex]\sqrt{\frac{5}{2} }i[/tex]
Those are the next 2 zeros. We found 4 so far, now we will find the last 2 in the second set of parenthesis above:
[tex]x^2-5=0[/tex] so
[tex]x^2=5[/tex] so
x = ±[tex]\sqrt{\frac{5}{2} }[/tex]
In summary, the 6 zeros are as follows:
x = [tex]\sqrt{5}i[/tex], -[tex]\sqrt{5}i[/tex], [tex]\sqrt{\frac{5}{2} }i[/tex], [tex]-\sqrt{\frac{5}{2} }i[/tex], [tex]\sqrt{\frac{5}{2} }[/tex], [tex]-\sqrt{\frac{5}{2} }[/tex]
