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

Question

25740 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T03:34:31

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

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 · 2018-06-08T03:49:44

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

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.
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 $f(x)$ .
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]}

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