Find roots of polynomial $$ p(x) = 24x^3+38x^2+19x+3 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 1 }{ 2 }\\[1 em]x_2 &= -\frac{ 1 }{ 3 }\\[1 em]x_3 &= -\frac{ 3 }{ 4 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = -\dfrac{ 1 }{ 2 } } $ is a root of polynomial $ 24x^3+38x^2+19x+3 $.
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}{ 3 } $, with factors of 1 and 3.
The leading coefficient is $ \color{red}{ 24 }$, with factors of 1, 2, 3, 4, 6, 8, 12 and 24.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 3 }}{\text{ factors of 24 }} = \pm \dfrac{\text{ ( 1, 3 ) }}{\text{ ( 1, 2, 3, 4, 6, 8, 12, 24 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2}\\[ 1 em] \pm \frac{ 1}{ 3} & \pm \frac{ 3}{ 3} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4}\\[ 1 em] \pm \frac{ 1}{ 6} & \pm \frac{ 3}{ 6} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 3}{ 8}\\[ 1 em] \pm \frac{ 1}{ 12} & \pm \frac{ 3}{ 12} ~~ \pm \frac{ 1}{ 24} \pm \frac{ 3}{ 24}\\[ 1 em] \end{aligned} $$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( -\dfrac{ 1 }{ 2 } \right) = 0 $ so $ x = -\dfrac{ 1 }{ 2 } $ 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}{ 2x+1 }$
$$ \frac{ 24x^3+38x^2+19x+3}{ 2x+1} = 12x^2+13x+3 $$Step 2:
The next rational root is $ x = -\dfrac{ 1 }{ 2 } $
$$ \frac{ 24x^3+38x^2+19x+3}{ 2x+1} = 12x^2+13x+3 $$Step 3:
The next rational root is $ x = -\dfrac{ 1 }{ 3 } $
$$ \frac{ 12x^2+13x+3}{ 3x+1} = 4x+3 $$Step 4:
To find the last zero, solve equation $ 4x+3 = 0 $
$$ \begin{aligned} 4x+3 & = 0 \\[1 em] 4 \cdot x & = -3 \\[1 em] x & = - \frac{ 3 }{ 4 } \end{aligned} $$