Find roots of polynomial $$ p(x) = -x^3+7x^2+4x-48 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= 6.4721\\[1 em]x_3 &= -2.4721 \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ -x^3+7x^2+4x-48 $.
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}{ 48 } $, with a single factor of 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
The leading coefficient is $ \color{red}{ 1 }$, with a single factor of 1.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 48 }}{\text{ factors of 1 }} = \pm \dfrac{\text{ ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 ) }}{\text{ ( 1 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 4}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 8}{ 1} \pm \frac{ 12}{ 1} \pm \frac{ 16}{ 1} \pm \frac{ 24}{ 1} \pm \frac{ 48}{ 1} ~~ \end{aligned} $$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( 3 \right) = 0 $ so $ x = 3 $ 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-3 }$
$$ \frac{ -x^3+7x^2+4x-48}{ x-3} = -x^2+4x+16 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ -x^3+7x^2+4x-48}{ x-3} = -x^2+4x+16 $$Step 3:
The solutions of $ -x^2+4x+16 = 0 $ are: $ x = 2-2 \sqrt{ 5 } ~ \text{and} ~ x = 2+2 \sqrt{ 5 }$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.