To execute so, you would need to element the denominator come \$(x^2+9)(x+3)(x-3)\$ and cancel the end \$(x-3)\$. If i wouldn"t have actually seen it being factored this way, it would"ve to be pretty difficult for me to aspect the denominator in 3 terms. Is there a basic way/method to factor instances like that, that would job-related in most cases?  One of my favored trick as soon as I watch a border for \$x o a\$ is to make \$x=y+a\$ and also work the limit for \$y o 0\$.

You are watching: Factor x^4-81

So, in her case\$\$fracx^4 - 3x^3x^4-81=frac27 y+27 y^2+9 y^3+y^4108 y+54 y^2+12 y^3+y^4 o frac 27108=frac 14\$\$ The an initial thing you try to evaluate the limit together \$x o 3\$ is come boldly shot and plug in \$x=3\$. You will an alert that both numerator and also denominator evaluate to \$0\$ there. This implies that performaing polynomial department by \$x-3\$ will succeed without remainder for both, whereafter you can cancel. Hopefully you understand that \$a^2-b^2=(a+b)(a-b)\$

We have the right to use this on the early equation to gain \$\$x^4-81=(x^2+9)(x^2-9)\$\$

We can then use the same method on the 2nd bracket to get \$\$x^2-9=(x+3)(x-3)\$\$

This gives us our last answer \$\$x^4-81=(x^2+9)(x+3)(x-3)\$\$ In this particular problem, as provided by others, the is more than likely simplest to variable using distinction of two squares twice.

More generally, though, if you space taking the limit of a quotient that polynomials together \$x o c\$ and get a \$frac00\$ form, then since \$c\$ is a zero of the top and of the bottom, the factor Theorem states that \$x-c\$ will certainly be a variable of both the top and bottom. This variable can be taken out, perhaps with man-made division, and also cancelled. (You don"t require to totally factor the numerator and also denominator necessarily, just gain them to the allude of gift able to cancel the \$x-c\$ factor.)

Write this as \$lim_x o 3 fracx^3(x-3)(x-3)(x+3)(x^2+9)=lim_x o 3 fracx^3(x+3)(x^2+9)=frac27108=frac14\$

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