Hayley Steele

2023-02-19

How to find all zeros of $f\left(x\right)={x}^{4}-{x}^{3}-20{x}^{2}$?

davz198888za

Beginner2023-02-20Added 9 answers

We have solution:

$\text{take out a common factor of}\phantom{\rule{1ex}{0ex}}{x}^{2}$

$f\left(x\right)={x}^{2}({x}^{2}-x-20)$

$\text{the factors of - 20 which sum to - 1 are - 5 and + 4}$

$\Rightarrow f\left(x\right)={x}^{2}(x-5)(x+4)$

$\text{to find zeros set}\phantom{\rule{1ex}{0ex}}f\left(x\right)=0$

$\Rightarrow {x}^{2}(x-5)(x+4)=0$

$\text{equate each factor to zero and solve for x}$

$x}^{2}=0\Rightarrow x=0\phantom{\rule{1ex}{0ex}}\text{with multiplicity 2$

$x-5=0\Rightarrow x=5$

$x+4=0\Rightarrow x=-4$

$\text{take out a common factor of}\phantom{\rule{1ex}{0ex}}{x}^{2}$

$f\left(x\right)={x}^{2}({x}^{2}-x-20)$

$\text{the factors of - 20 which sum to - 1 are - 5 and + 4}$

$\Rightarrow f\left(x\right)={x}^{2}(x-5)(x+4)$

$\text{to find zeros set}\phantom{\rule{1ex}{0ex}}f\left(x\right)=0$

$\Rightarrow {x}^{2}(x-5)(x+4)=0$

$\text{equate each factor to zero and solve for x}$

$x}^{2}=0\Rightarrow x=0\phantom{\rule{1ex}{0ex}}\text{with multiplicity 2$

$x-5=0\Rightarrow x=5$

$x+4=0\Rightarrow x=-4$