Question

Please solve and show steps

Answer 1

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:

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

$x=\sqrt{-5}$

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 = ±$\sqrt{\frac{5}{2} }i$

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 = ±$\sqrt{\frac{5}{2} }$

In summary, the 6 zeros are as follows:

x = $\sqrt{5}i$, -$\sqrt{5}i$, $\sqrt{\frac{5}{2} }i$, $-\sqrt{\frac{5}{2} }i$, $\sqrt{\frac{5}{2} }$, $-\sqrt{\frac{5}{2} }$