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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?

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  • $\begingroup$ Think again about $(x+5)(x+5)$. $\endgroup$ – Gerry Myerson Feb 16 '14 at 8:28
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    $\begingroup$ $x^2 + 25 \not= (x+5)(x+5)$ $\endgroup$ – Shaurya Gupta Feb 16 '14 at 8:35
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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.

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  • $\begingroup$ Thank you for the answers so, (X^2 + any number) always have no solution ? $\endgroup$ – Helena Feb 16 '14 at 9:04
  • $\begingroup$ @Helena : if 'any number' is (strictly) positive, yes. (for real numbers anyway) $\endgroup$ – imj Feb 16 '14 at 9:12
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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}$

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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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