x-Intercepts of Polynomial Functions Practice

We want to find the zeros of these polynomials: 

1. $p(x)=3x^3-3x^2-18x$
2. $p(x)=2x^3-x^2-8x+4$
3. $p(x)=(x^2-9)(x^2+x-2)$
4. $p(x)=(2x^2-9x+7)(x-2)$

Plot all the zeros ($x$-intercepts) of the polynomial in the graph.

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-graphs/x2ec2f6f830c9fb89:poly-zeros/e/find-the-zeros-of-polynomials
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. First, we need to factor the polynomial entirely.

   Then, we can find the zeros by looking for all the $x$-values that make any of the factors equal to zero.

   Factoring
   

   As a first step, we can take out a common factor:

   $\begin{aligned}p(x)&=3x^3-3x^2-18x\\\\&=3x(x^2-x-6)\end{aligned}$
   

   Now we can further factor $(x^2-x-6)$:

   $\begin{aligned}&\phantom{{}={}}3x {(x^2-x-6)}\\\\&=3x {(x-3)(x+2)}\end{aligned}$

   Finding the zeros
   

   So this is the factored form of $p$:
   

   $p(x)=3x(x-3)(x+2)$

   This table contains all the factors of $p$ and the $x$-values that make them equal to zero.

   Factor	Zero
   $3x$	$x=0$
   $(x-3)$	$x=3$
   $(x+2)$	$x=-2$

   
   We found that the zeros of $p$ are $x=0$, $x=3$, and $x=-2$, which means the graph has $x$-intercepts at $(0,0)$, $(3,0)$, and $(-2,0)$.

   
   

   ---

2. First, we need to factor the polynomial entirely.

   Then, we can find the zeros by looking for all the $x$-values that make any of the factors equal to zero.

   Factoring

   As a first step, we can factor the polynomial using grouping:

   $\begin{aligned}p(x)&=2x^3-x^2-8x+4\\\\&=x^2(2x-1)-4(2x-1)\\\\&=(x^2-4)(2x-1)\end{aligned}$
   

   Now we can further factor $(x^2-4)$ using the difference of squares pattern:

   $\begin{aligned}&\phantom{{}={}} {(x^2-4)}(2x-1)\\\\&= {(x+2)(x-2)}(2x-1)\end{aligned}$

   Finding the zeros

   So this is the factored form of $p$:
   

   $p(x)=(x+2)(x-2)(2x-1)$

   This table contains all the factors of $p$ and the $x$-values that make them equal to zero.

   Factor	Zero
   $(x+2)$	$x=-2$
   $(x-2)$	$x=2$
   $(2x-1)$	$x=\dfrac{1}2$

   
   

   We found that the zeros of $p$ are $x=-2$, $x=2$, and $x=\dfrac{1}2$, which means the graph has $x$-intercepts at $(-2,0)$, $(2,0)$, and $\left(\dfrac{1}2,0\right)$.

   
   

   ---

3. First, we need to factor the polynomial entirely.

   Then, we can find the zeros by looking for all the $x$-values that make any of the factors equal to zero.

   Factoring

   $p$ is already partially factored. We can further factor $(x^2-9)$ using the difference of squares pattern:

   $\begin{aligned}&\phantom{{}={}} {(x^2-9)}(x^2+x-2)\\\\&= {(x+3)(x-3)}(x^2+x-2)\end{aligned}$
   

   Now we can further factor $(x^2+x-2)$:

   $\begin{aligned}&\phantom{{}={}}(x+3)(x-3) {(x^2+x-2)}\\\\&=(x+3)(x-3) {(x+2)(x-1)}\end{aligned}$

   Finding the zeros

   So this is the factored form of $p$:
   

   $p(x)=(x+3)(x-3)(x+2)(x-1)$

   This table contains all the factors of $p$ and the $x$-values that make them equal to zero.

   Factor	Zero
   $(x+3)$	$x=-3$
   $(x-3)$	$x=3$
   $(x+2)$	$x=-2$
   $(x-1)$	$x=1$

   
   We found that the zeros of $p$ are $x=-3$, $x=3$, $x=-2$, and $x=1$, which means the graph has $x$-intercepts at $\left(-3,0\right)$, $\left(3,0\right)$, $\left(-2,0\right)$, and $\left(1,0\right)$.

   
   

   ---

4. First, we need to factor the polynomial entirely.

   Then, we can find the zeros by looking for all the $x$-values that make any of the factors equal to zero.

   Factoring

   $p$ is already partially factored. We can further factor $(2x^2-9x+7)$ using the grouping method:

   $\begin{aligned}p(x)&=(2x^2-9x+7)(x-2)\\\\&=(2x^2-2x-7x+7)(x-2)\\\\&=\big(2x(x-1)-7(x-1)\big)(x-2)\\\\&=(2x-7)(x-1)(x-2)\end{aligned}$

   Finding the zeros

   So this is the factored form of $p$:
   

   $p(x)=(2x-7)(x-1)(x-2)$

   This table contains all the factors of $p$ and the $x$-values that make them equal to zero.

   Factor	Zero
   $(2x-7)$	$x=\dfrac{7}2$
   $(x-1)$	$x=1$
   $(x-2)$	$x=2$

   
   

   We found that the zeros of $p$ are $x=\dfrac{7}2$, $x=1$, and $x=2$, which means the graph has $x$-intercepts at  $\left(\dfrac{7}2,0\right)$, $\left(1,0\right)$, and $\left(2,0\right)$.