How do you find all the zeros of $f(x)=x^3-x^2+49x-49$ ?

Question

2548 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-19T23:07:30

$f(x)$ has zeros $+-7i$ and $1$

This cubic factors by grouping, then by using the difference of squares identity:

$a^2=b^2=(a-b)(a+b)$

with $a=x$ and $b=7i$ as follows:

$f(x) = x^3-x^2+49x-49$

$=(x^3-x^2)+(49x-49)$

$=x^2(x-1)+49(x-1)$

$=(x^2+49)(x-1)$

$=(x^2-(7i)^2)(x-1)$

$=(x-7i)(x+7i)(x-1)$

Hence $f(x)$ has zeros $+-7i$ and $1$

How do you find all the zeros of f(x)=x^3-x^2+49x-49f(x)=x^3-x^2+49x-49?