Find the x and y intercepts of $y=x^4-8x^2$ The question asks find the co-ordinates of the points where the graph $y= f(x) $crosses the axes. 
I know to change it to $y=x^4-8x^2$ but then how do I factorise that?
 A: $y=x^4-8x^2$ crosses the axes when $x=0$ or $y=0$:


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*$x=0$ leads to $y=0^4-8\cdot0^2=0$, so $y=x^4-8x^2$ crosses the axes in $(0,0)$.

*$y=0$ leads to $x^4-8x^2=0$. Can you solve that equation?

A: It's difficult to know how much detail to go in to without knowing what you've already tried, and how confident you are with similar questions, so I will give hints only.
You want to factorise $$x^4-8x^2.$$ Both of these terms have a factor of $x^2$, so it's easy to take that out, $$x^2(x^2-8).$$ So now to factorise further, you need to split $(x^2-8)$ in to two factors. If this is hard, try factorising $(x^2-16)$ for example (you may have done this one or similar before in your course).
If you're still not sure, try multiplying a couple of things together and see what you get. For example, $(x+1)(x+1)=x^2+2x+1$. What about $(x-1)(x-1)$ or $(x+1)(x-1)$?
Other techniques you might use include:


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*actually try solving the equation $x^2-8=0$ using the quadratic formula, or completing the square.

*try some numbers for x as a guess.


From what you say, it sounds like you know how to find the interception points after factorising. Just in case, finding x interception points means finding what values of $x$ we have if $y=0$, and y interception points is about finding what values of $y$ we have if $x=0$.
