Hello, Scotland!
$\displaystyle f(x)\:=\: x^3 x^25x  3$
a) i) Show that $\displaystyle (x+1)$ is a factor of $\displaystyle f(x)$
. .ii) Factorise $\displaystyle f(x)$ fully. We can use the Remainder/Factor Theorem.
. . $\displaystyle f(\text{}1) \:=\\text{}1)^3  (\text{}1)^2  5(\text{}1)  3 \;=\;0$
Since $\displaystyle f(\text{}1) = 0$, then $\displaystyle (x+1)$ is a factor of $\displaystyle f(x).$
Then we have: .$\displaystyle f(x) \;=\;(x+1)(x^22x3) \;=\;(x+1)^2(x3)$
b) One of the turning points of the graph of $\displaystyle f(x)$
lies on the xaxis. .Find the coordinates of this turning point. We can answer this without Calculus . . .
In part (a), we found that: .$\displaystyle f(x) \:=\x+1)^2(x3)$
The curve has xintercepts at $\displaystyle (1,\,0)\text{ and }(3,\,0)$.
The intercept $\displaystyle (1,\,0)$ has multiplicity 2.
. . Hence, the curve is tangent to the xaxis at $\displaystyle x = 1.$
Therefore, there is a turning point at $\displaystyle (1,\,0).$
The graph looks like this: Code:

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o+o
* 1 *  * 3
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