What is the $x$ intercept of $y=(x-2)(x^2+25) $?
To find $x$ intercept:$ 0=(x-2)(x^2+25) $
I tried $ 0=(x-2)(x+5)(x+5)$ in which the $X$ intercepts are $2,-5$ and $-5$. Is this correct?
$\endgroup$ 23 Answers
$\begingroup$no. $(x+5)(x+5)\neq (x^2+25)$. Assuming you are solving for real $x$, $x^2+25=0$ has no solutions, as the square of a real is always non-negative.
$\endgroup$ 2 $\begingroup$You are correct upto following
To find $x$ intercept : $ 0=(x-2)(x^2+25) $
Now notice that $(x^2+25) > 0 $ $ \forall x \in \mathbb{R}$ and he factorization which you did is wrong.
Thus $ 0=(x-2)(x^2+25) $ has only one root namely $ x=2$ hence that is the intercept.
Note : $x^2 > 0 $ $ \forall x \in \mathbb{R}$and $25$ being a positive integer we get that $(x^2+25) > 0 $ $ \forall x \in \mathbb{R}$
$\endgroup$ $\begingroup$Your answer is incorrect. $(x+5)(x+5)=x^2+10x+25 \neq x^2+25$. One intercept is $2$, and it is the only x-intercept that you can graph with real numbers. The FULL factorization of the cubic is $(x-2)(x+5i)(x-5i)$, where $i=\sqrt{-1}$. $i$ is the basic unit for representing complex numbers; look it up. To graph the other intercepts you have to use the complex plane.
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