Find roots of polynomial $$ p(x) = -x^5+3x^2+4x^3-28x^2+37x-15 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1\\[1 em]x_2 &= 0.7235\\[1 em]x_3 &= -3.706\\[1 em]x_4 &= 0.9912+2.1475i\\[1 em]x_5 &= 0.9912-2.1475i \end{aligned} $$Explanation
Step 1:
Combine like terms:
$$ -x^5+ \color{blue}{3x^2} +4x^3 \color{blue}{-28x^2} +37x-15 = -x^5+4x^3 \color{blue}{-25x^2} +37x-15 $$Step 2:
Use rational root test to find out that the $ \color{blue}{ x = 1 } $ is a root of polynomial $ -x^5+4x^3-25x^2+37x-15 $.
The Rational Root Theorem tells us that if the polynomial has a rational zero then it must be a fraction $ \dfrac{ \color{blue}{p}}{ \color{red}{q} } $, where $ p $ is a factor of the constant term and $ q $ is a factor of the leading coefficient.
The constant term is $ \color{blue}{ 15 } $, with a single factor of 1, 3, 5 and 15.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 15 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 3, 5, 15 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 5}{ 1} \pm \frac{ 15}{ 1} ~~ \end{aligned} $$Substitute the possible roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.
We can see that $ p\left( 1 \right) = 0 $ so $ x = 1 $ is a root of a polynomial $ p(x) $.
To find remaining zeros we use Factor Theorem. This theorem states that if $ \dfrac{p}{q} $ is root of the polynomial then the polynomial can be divided by $ \color{blue}{qx − p} $. In this example we divide polynomial $ p $ by $ \color{blue}{ x-1 }$
$$ \frac{ -x^5+4x^3-25x^2+37x-15}{ x-1} = -x^4-x^3+3x^2-22x+15 $$Step 3:
The next rational root is $ x = 1 $
$$ \frac{ -x^5+4x^3-25x^2+37x-15}{ x-1} = -x^4-x^3+3x^2-22x+15 $$Step 4:
Polynomial $ -x^4-x^3+3x^2-22x+15 $ has no rational roots that can be found using Rational Root Test, so the roots were found using quartic formulas.