Question

$${ x }^{ 3 } +5x { y }^{ 2 } +3 { x }^{ 4 } { y }^{ 3 } ThisIsPollynomial$$

Answer

$$x^3+5*x*y^2-3*Th*sIsPo*x^4*y^4*l^3*n*o*m*a$$

Solution


Regroup terms.
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{3}ylllnomaTh\imath sIsPo\imath \]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{3+1}{l}^{1+1+1}nomaTh\imath sIsPo\imath \]
Simplify  \(3+1\)  to  \(4\).
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{4}{l}^{1+1+1}nomaTh\imath sIsPo\imath \]
Simplify  \(1+1\)  to  \(2\).
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{4}{l}^{2+1}nomaTh\imath sIsPo\imath \]
Simplify  \(2+1\)  to  \(3\).
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{4}{l}^{3}nomaTh\imath sIsPo\imath \]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{4}{l}^{3}nomaTh{\imath }^{2}sIsPo\]
Use Square Rule: \({i}^{2}=-1\).
\[{x}^{3}+5x{y}^{2}+3{x}^{4}{y}^{4}{l}^{3}nomaTh\times -1\times sIsPo\]
Simplify  \(3{x}^{4}{y}^{4}{l}^{3}nomaTh\times -1\times sIsPo\)  to  \(-3{x}^{4}{y}^{4}{l}^{3}nomaThsIsPo\).
\[{x}^{3}+5x{y}^{2}-3{x}^{4}{y}^{4}{l}^{3}nomaThsIsPo\]
Regroup terms.
\[{x}^{3}+5x{y}^{2}-3ThsIsPo{x}^{4}{y}^{4}{l}^{3}noma\]