Question

Please solve and show steps

Answer 1

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:

`(4x^6+20x^4)-(25x^2-125)` then

`4x^4(x^2+5)-25(x^2+5)` then

`(x^2+5)(4x^4-25)=0`

We will factor each set of parenthesis now to get all the zeros. For the first set of parenthesis:

`x^2+5=0` so

`x^2=-5` so

`x=√(-5)`

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 = ±
`\sqrt{(5)/(2) }i`

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:

`x^2-5=0` so

`x^2=5` so

x = ±
`\sqrt{(5)/(2) }`

In summary, the 6 zeros are as follows:

x =
`√(5)i`, -
`√(5)i`,
`\sqrt{(5)/(2) }i`,
`-\sqrt{(5)/(2) }i`,
`\sqrt{(5)/(2) }`,
`-\sqrt{(5)/(2) }`