# How do you find the zeros of the polynomial function f(x) = x^3 + x^2 -42x?

Jun 8, 2018

$x = 0 , x = 6$, and $x = - 7$

#### Explanation:

To start, every term has an $x$ in common, so we can factor that out. Doing this, we get

$x \textcolor{b l u e}{\left({x}^{2} + x - 42\right)} = 0$

What I have in blue can be factored by a thought experiment. What two numbers have a sum of $1$ and a product of $- 42$?

After some trial and error, we arrive at $- 6$ and $7$. Thus, this business can be factored as

$x \left(x - 6\right) \left(x + 7\right) = 0$

We can leverage the zero product property here, setting all three terms equal to zero. As our zeroes, we get

$x = 0 , x = 6$, and $x = - 7$

Hope this helps!

Jun 8, 2018

$x = 0$,$x = 6$ and $x = - 7$ are the roots of$f \left(x\right)$

#### Explanation:

1. METHOD 1 : Try and Error Method
We can find out the zeros (roots) of the polynomial. by the 'Try and Error' method. In this method, we prefer to substitute the integer in given polynomial and wait for zero.
But, this is not fair and good method.
2. METHOD 2 : Observation Method
This method is based on observation. In some questions we can apply this. It is important to click this method.
By observation, it is cleared that $x = 0$ is one of the roots.
$\therefore$ $f \left(x\right) = x \left[{x}^{2} + x - 42\right]$
Now, we have to find out the roots of the quadratic equation ${x}^{2} + x - 42$. It is easier one.
$\therefore$ $f \left(x\right) = x \left[{x}^{2} + x - 42\right]$
$\therefore$ $f \left(x\right) = x \left[{x}^{2} + 7 x - 6 x - 42\right]$
$\therefore$ $f \left(x\right) = x \left[x \left(x + 7\right) - 6 \left(x + 7\right)\right]$
$\therefore$ $f \left(x\right) = x \left(x - 6\right) \left(x + 7\right)$
Hence, $x = 0$,$x = 6$ and $x = - 7$ are the roots of$f \left(x\right)$.
3. METHOD 3 : Graph Method
We can draw rough sketch of the graph of $f \left(x\right)$ with the help of Calculus.
graph{x^3+x^2-42x [-8.89, 8.88, -4.444, 4.445]}