Find roots of polynomial $$ p(x) = 16x^4-40x^3+x^2+39x-18 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 2\\[1 em]x_2 &= -1\\[1 em]x_3 &= \frac{ 3 }{ 4 }\\[1 em]x_4 &= \frac{ 3 }{ 4 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 2 } $ is a root of polynomial $ 16x^4-40x^3+x^2+39x-18 $.
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}{ 18 } $, with factors of 1, 2, 3, 6, 9 and 18.
The leading coefficient is $ \color{red}{ 16 }$, with factors of 1, 2, 4, 8 and 16.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 18 }}{\text{ factors of 16 }} = \pm \dfrac{\text{ ( 1, 2, 3, 6, 9, 18 ) }}{\text{ ( 1, 2, 4, 8, 16 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 2}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 6}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 18}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 2}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 6}{ 2} \pm \frac{ 9}{ 2} \pm \frac{ 18}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 2}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 6}{ 4} \pm \frac{ 9}{ 4} \pm \frac{ 18}{ 4} ~~ \pm \frac{ 1}{ 8} \pm \frac{ 2}{ 8} \pm \frac{ 3}{ 8} \pm \frac{ 6}{ 8} \pm \frac{ 9}{ 8} \pm \frac{ 18}{ 8} ~~ \pm \frac{ 1}{ 16} \pm \frac{ 2}{ 16} \pm \frac{ 3}{ 16} \pm \frac{ 6}{ 16} \pm \frac{ 9}{ 16} \pm \frac{ 18}{ 16} ~~ \end{aligned} $$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( 2 \right) = 0 $ so $ x = 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}{ x-2 }$
$$ \frac{ 16x^4-40x^3+x^2+39x-18}{ x-2} = 16x^3-8x^2-15x+9 $$Step 2:
The next rational root is $ x = 2 $
$$ \frac{ 16x^4-40x^3+x^2+39x-18}{ x-2} = 16x^3-8x^2-15x+9 $$Step 3:
The next rational root is $ x = -1 $
$$ \frac{ 16x^3-8x^2-15x+9}{ x+1} = 16x^2-24x+9 $$Step 4:
The next rational root is $ x = \dfrac{ 3 }{ 4 } $
$$ \frac{ 16x^2-24x+9}{ 4x-3} = 4x-3 $$Step 5:
To find the last zero, solve equation $ 4x-3 = 0 $
$$ \begin{aligned} 4x-3 & = 0 \\[1 em] 4 \cdot x & = 3 \\[1 em] x & = \frac{ 3 }{ 4 } \end{aligned} $$