Simplify \(12y-3{y}^{2}-18{y}^{3}-3-32y+57{y}^{2}+5-18{y}^{3}\) to \(-20y+54{y}^{2}-36{y}^{3}+2\).
\[-20y+54{y}^{2}-36{y}^{3}+2=0\]
Factor out the common term \(2\).
\[-2(10y-27{y}^{2}+18{y}^{3}-1)=0\]
Factor \(10y-27{y}^{2}+18{y}^{3}-1\) using Polynomial Division.
Factor the following.
\[10y-27{y}^{2}+18{y}^{3}-1\]
First, find all factors of the constant term 1.
\[1\]
Try each factor above using the Remainder Theorem.
Substitute 1 into y. Since the result is 0, y-1 is a factor..\(10\times 1-27\times {1}^{2}+18\times {1}^{3}-1 = 0\)
\[y-1\]
Polynomial Division: Divide \(10y-27{y}^{2}+18{y}^{3}-1\) by \(y-1\).
\[18y^2\]
\[-9y\]
\[1\]
\[y-1\]
\[18y^3\]
\[-27y^2\]
\[10y\]
\[-1\]
\[18y^3\]
\[-18y^2\]
\[-9y^2\]
\[10y\]
\[-1\]
\[-9y^2\]
\[9y\]
\[y\]
\[-1\]
\[y\]
\[-1\]
\[\]
Rewrite the expression using the above.
\[18{y}^{2}-9y+1\]
\[-2(18{y}^{2}-9y+1)(y-1)=0\]
Split the second term in \(18{y}^{2}-9y+1\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[18\times 1=18\]
Ask: Which two numbers add up to \(-9\) and multiply to \(18\)?
\(-3\) and \(-6\)
Split \(-9y\) as the sum of \(-3y\) and \(-6y\).
\[18{y}^{2}-3y-6y+1\]
\[-2(18{y}^{2}-3y-6y+1)(y-1)=0\]
Factor out common terms in the first two terms, then in the last two terms.
\[-2(3y(6y-1)-(6y-1))(y-1)=0\]
Factor out the common term \(6y-1\).
\[-2(6y-1)(3y-1)(y-1)=0\]
Divide both sides by \(-2\).
\[(6y-1)(3y-1)(y-1)=0\]
Solve for \(y\).
Ask: When will \((6y-1)(3y-1)(y-1)\) equal zero?
When \(6y-1=0\), \(3y-1=0\), or \(y-1=0\)
Solve each of the 3 equations above.
\[y=\frac{1}{6},\frac{1}{3},1\]
\[y=\frac{1}{6},\frac{1}{3},1\]
Check solution
When \(y=\frac{1}{3}\), the original equation \(\frac{20y}{36{y}^{2}-4}-\frac{2y-3}{2-6y}=\frac{5-2y}{6y+2}\) does not hold true.We will drop \(y=\frac{1}{3}\) from the solution set.
