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9y^{4}-7y^{2}-16=0
To factor the expression, solve the equation where it equals to 0.
±\frac{16}{9},±\frac{16}{3},±16,±\frac{8}{9},±\frac{8}{3},±8,±\frac{4}{9},±\frac{4}{3},±4,±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{9},±\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 -16 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.
y=\frac{4}{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.
3y^{3}+4y^{2}+3y+4=0
By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 9y^{4}-7y^{2}-16 by 3\left(y-\frac{4}{3}\right)=3y-4 to get 3y^{3}+4y^{2}+3y+4. To factor the result, solve the equation where it equals to 0.
±\frac{4}{3},±4,±\frac{2}{3},±2,±\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 4 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.
y=-\frac{4}{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.
y^{2}+1=0
By Factor theorem, y-k is a factor of the polynomial for each root k. Divide 3y^{3}+4y^{2}+3y+4 by 3\left(y+\frac{4}{3}\right)=3y+4 to get y^{2}+1. To factor the result, solve the equation where it equals to 0.
y=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}
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 1 for a, 0 for b, and 1 for c in the quadratic formula.
y=\frac{0±\sqrt{-4}}{2}
Do the calculations.
y^{2}+1
Polynomial y^{2}+1 is not factored since it does not have any rational roots.
\left(3y-4\right)\left(3y+4\right)\left(y^{2}+1\right)
Rewrite the factored expression using the obtained roots.