Question

Consider the function f(x) = x^3 - 7x^2 - 2x + 14

Answer 1

Answer:

There are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14$

To get the zeros of $f(x)$, set $y$ or $f(x) =0$

so

$0=x^3-7x^2-2x+14$

as

$x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)$

so

$\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0$

Using the zero factor principle:

$\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0$

solving

$x-7=0$

• $x=7$

$x+\sqrt{2}=0$

• $x=-\sqrt{2}$

$x-\sqrt{2}=0$

• $x=\sqrt{2}$

Therefore, there are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$