Question

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

Answer 1

`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

`xcolor(blue)((x^2+x-42))=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(x-6)(x+7)=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!

Answer 2

`x=0`,`x=6` and `x=-7` are the roots of`f(x)`

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.
   `:.` `f(x)=x[x^2+x-42]`
   Now, we have to find out the roots of the quadratic equation `x^2+x-42`. It is easier one.
   `:.` `f(x)=x[x^2+x-42]`
   `:.` `f(x)=x[x^2+7x-6x-42]`
   `:.` `f(x)=x[x(x+7)-6(x+7)]`
   `:.` `f(x)=x(x-6)(x+7)`
   Hence, `x=0`,`x=6` and `x=-7` are the roots of`f(x)`.
3. METHOD 3 : Graph Method
   We can draw rough sketch of the graph of `f(x)` with the help of Calculus.
   graph{x^3+x^2-42x [-8.89, 8.88, -4.444, 4.445]}