Find roots of polynomial $$ p(x) = 4x^4+8x^3-17x^2-12x+9 $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= -1\\[1 em]x_2 &= -3\\[1 em]x_3 &= \frac{ 1 }{ 2 }\\[1 em]x_4 &= \frac{ 3 }{ 2 } \end{aligned} $$

Step 1:

Use rational root test to find out that the $ \color{blue}{ x = -1 } $ is a root of polynomial $ 4x^4+8x^3-17x^2-12x+9 $.

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}{ 9 } $, with factors of 1, 3 and 9.

The leading coefficient is $ \color{red}{ 4 }$, with factors of 1, 2 and 4.

The POSSIBLE zeroes are:

$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 9 }}{\text{ factors of 4 }} = \pm \dfrac{\text{ ( 1, 3, 9 ) }}{\text{ ( 1, 2, 4 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 9}{ 1} ~~ \pm \frac{ 1}{ 2} \pm \frac{ 3}{ 2} \pm \frac{ 9}{ 2} ~~ \pm \frac{ 1}{ 4} \pm \frac{ 3}{ 4} \pm \frac{ 9}{ 4} ~~ \end{aligned} $$

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 }$

$$ \frac{ 4x^4+8x^3-17x^2-12x+9}{ x+1} = 4x^3+4x^2-21x+9 $$

Step 2:

The next rational root is $ x = -1 $

$$ \frac{ 4x^4+8x^3-17x^2-12x+9}{ x+1} = 4x^3+4x^2-21x+9 $$

Step 3:

The next rational root is $ x = -3 $

$$ \frac{ 4x^3+4x^2-21x+9}{ x+3} = 4x^2-8x+3 $$

Step 4:

The next rational root is $ x = \dfrac{ 1 }{ 2 } $

$$ \frac{ 4x^2-8x+3}{ 2x-1} = 2x-3 $$

Step 5:

To find the last zero, solve equation $ 2x-3 = 0 $

$$ \begin{aligned} 2x-3 & = 0 \\[1 em] 2 \cdot x & = 3 \\[1 em] x & = \frac{ 3 }{ 2 } \end{aligned} $$

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