Find roots of polynomial $$ p(x) = 5x^3-7x^2-45x+63 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3\\[1 em]x_2 &= -3\\[1 em]x_3 &= \frac{ 7 }{ 5 } \end{aligned} $$Explanation
Step 1:
Use rational root test to find out that the $ \color{blue}{ x = 3 } $ is a root of polynomial $ 5x^3-7x^2-45x+63 $.
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}{ 63 } $, with factors of 1, 3, 7, 9, 21 and 63.
The leading coefficient is $ \color{red}{ 5 }$, with factors of 1 and 5.
The POSSIBLE zeroes are:
$$ \begin{aligned} \dfrac{\color{blue}{p}}{\color{red}{q}} = & \dfrac{ \text{ factors of 63 }}{\text{ factors of 5 }} = \pm \dfrac{\text{ ( 1, 3, 7, 9, 21, 63 ) }}{\text{ ( 1, 5 ) }} = \\[1 em] = & \pm \frac{ 1}{ 1} \pm \frac{ 3}{ 1} \pm \frac{ 7}{ 1} \pm \frac{ 9}{ 1} \pm \frac{ 21}{ 1} \pm \frac{ 63}{ 1} ~~ \pm \frac{ 1}{ 5} \pm \frac{ 3}{ 5} \pm \frac{ 7}{ 5} \pm \frac{ 9}{ 5} \pm \frac{ 21}{ 5} \pm \frac{ 63}{ 5} ~~ \end{aligned} $$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 }$
$$ \frac{ 5x^3-7x^2-45x+63}{ x-3} = 5x^2+8x-21 $$Step 2:
The next rational root is $ x = 3 $
$$ \frac{ 5x^3-7x^2-45x+63}{ x-3} = 5x^2+8x-21 $$Step 3:
The next rational root is $ x = -3 $
$$ \frac{ 5x^2+8x-21}{ x+3} = 5x-7 $$Step 4:
To find the last zero, solve equation $ 5x-7 = 0 $
$$ \begin{aligned} 5x-7 & = 0 \\[1 em] 5 \cdot x & = 7 \\[1 em] x & = \frac{ 7 }{ 5 } \end{aligned} $$