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45x^{4}+4x^{2}-1=0
To factor the expression, solve the equation where it equals to 0.
±\frac{1}{45},±\frac{1}{15},±\frac{1}{9},±\frac{1}{5},±\frac{1}{3},±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 45. List all candidates \frac{p}{q}.
x=\frac{1}{3}
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
15x^{3}+5x^{2}+3x+1=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 45x^{4}+4x^{2}-1 by 3\left(x-\frac{1}{3}\right)=3x-1 to get 15x^{3}+5x^{2}+3x+1. To factor the result, solve the equation where it equals to 0.
±\frac{1}{15},±\frac{1}{5},±\frac{1}{3},±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 15. List all candidates \frac{p}{q}.
x=-\frac{1}{3}
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
5x^{2}+1=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 15x^{3}+5x^{2}+3x+1 by 3\left(x+\frac{1}{3}\right)=3x+1 to get 5x^{2}+1. To factor the result, solve the equation where it equals to 0.
x=\frac{0±\sqrt{0^{2}-4\times 5\times 1}}{2\times 5}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 5 for a, 0 for b, and 1 for c in the quadratic formula.
x=\frac{0±\sqrt{-20}}{10}
Do the calculations.
5x^{2}+1
Polynomial 5x^{2}+1 is not factored since it does not have any rational roots.
\left(3x-1\right)\left(3x+1\right)\left(5x^{2}+1\right)
Rewrite the factored expression using the obtained roots.