so far I only got 2. which I got (x-1)(x^4+9x^2+20) but I am stuck otherwise if you can please show what next steps are I greatly appreciate it. thankyou
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There aren't any rational solutions to 1. So the best you can do is look at the graph and estimate the only real root at x ≈ 0.7504970613289412
The other four roots are complex.
(x-1)(x^4+9x^2+20) = x^5 - x^4 + 9x^3 - 9x^2 + 20x - 20, which is not what you started with, so (x-1) is not to the point.
On the other hand, supposing you miscopied the problem, and (x-1)(x^4+9x^2+20) was actually correct. In that case, make the simplifying substitution
y = x^2, y^2 = x^4:
y^2 + 9y + 20
Factor:
(y + 4) (y + 5)
Undo the substitution:
(x^2 + 4) (x^2 + 5)
Neither of these quadratics factors in real numbers, but the quadratic formula can be used to solve each one:
x = ± 2i where i^2 = -1
and
x = ± i √ 5
For the second problem, look at the graph or use the rational roots theorem to find the root at x = -2.
Use the factor theorem to convert this root into a factor:
(x + 2)
Divide by the known factor:
(x^5 + 4x^3 + 8x^2 + 32) / (x + 2) = x^4 - 2x^3 + 8x^2 - 8x + 16
The reduced quartic factors into:
(x^2 + 4) (x^2 - 2x + 4)
Neither of these quadratics factor in real numbers, but you can use the quadratic formula on each of them separately to find:
x = ± 2i
and
x = 1 ± i √ 3
The other four roots are complex.
(x-1)(x^4+9x^2+20) = x^5 - x^4 + 9x^3 - 9x^2 + 20x - 20, which is not what you started with, so (x-1) is not to the point.
On the other hand, supposing you miscopied the problem, and (x-1)(x^4+9x^2+20) was actually correct. In that case, make the simplifying substitution
y = x^2, y^2 = x^4:
y^2 + 9y + 20
Factor:
(y + 4) (y + 5)
Undo the substitution:
(x^2 + 4) (x^2 + 5)
Neither of these quadratics factors in real numbers, but the quadratic formula can be used to solve each one:
x = ± 2i where i^2 = -1
and
x = ± i √ 5
For the second problem, look at the graph or use the rational roots theorem to find the root at x = -2.
Use the factor theorem to convert this root into a factor:
(x + 2)
Divide by the known factor:
(x^5 + 4x^3 + 8x^2 + 32) / (x + 2) = x^4 - 2x^3 + 8x^2 - 8x + 16
The reduced quartic factors into:
(x^2 + 4) (x^2 - 2x + 4)
Neither of these quadratics factor in real numbers, but you can use the quadratic formula on each of them separately to find:
x = ± 2i
and
x = 1 ± i √ 3