# Finding X intercept of a cubic equation?

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?

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

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.

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

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 $(x^2+25) > 0$ $\forall x \in \mathbb{R}$

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.