Article: When solving an equation with binomials, especially complex binomials, it can seem like there is no way everything will match. For example, try to solve 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y}. One way to solve it, especially with exponents, is to factor first.   Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y}  Remember that binomials must only have two terms. If there are more than two terms you can learn to solve polynomials instead. This whole strategy relies on one of the most basic facts of math: anything multiplied by zero must equal zero. So if you equation equals zero, then one of your factored terms must equal zero! To get started, add and subtract so one side equals zero.   Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y}   Set to Zero: 5y−2y2+3y=−3y+3y{\displaystyle 5y-2y^{2}+3y=-3y+3y} 8y−2y2=0{\displaystyle 8y-2y^{2}=0} At this point, you can pretend the other side doesn't exist for a step. Just find the greatest common factor, divided it out, and then create your factored expression.   Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y}   Set to Zero: 8y−2y2=0{\displaystyle 8y-2y^{2}=0}   Factor: 2y(4−y)=0{\displaystyle 2y(4-y)=0} In the practice problem you are multiplying 2y by 4 - y, and it must equal zero. Since anything multiplied by zero equals zero, this means either 2y or 4 - y must be 0. Create two separate equations to figure out what y must be for either side to equal zero.   Practice Problem: 5y−2y2=−3y{\displaystyle 5y-2y^{2}=-3y}   Set to Zero: 8y−2y2+3y=0{\displaystyle 8y-2y^{2}+3y=0}   Factor: 2y(4−y)=0{\displaystyle 2y(4-y)=0}   Set both parts to 0:  2y=0{\displaystyle 2y=0} 4−y=0{\displaystyle 4-y=0} You might have one answer, or more than one. Remember, only one side has to equal zero, so you might get a few different values of y that solve the same equation. For the end of the practice problem:   2y=0{\displaystyle 2y=0}  2y2=02{\displaystyle {\frac {2y}{2}}={\frac {0}{2}}} y = 0    4−y=0{\displaystyle 4-y=0}  4−y+y=0+y{\displaystyle 4-y+y=0+y} y = 4 If you got the right values for y then you should be able to use them to solve the equation. It is simple as trying each value of y in place of the variable, as shown. Since the answer were y = 0 and y = 4:   5(0)−2(0)2=−3(0){\displaystyle 5(0)-2(0)^{2}=-3(0)}  0+0=0{\displaystyle 0+0=0}  0=0{\displaystyle 0=0} This answer is correct     5(4)−2(4)2=−3(4){\displaystyle 5(4)-2(4)^{2}=-3(4)}  20−32=−12{\displaystyle 20-32=-12}  −12=−12{\displaystyle -12=-12} This answer is also correct.
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Use factoring to simplify equations and make them easier to solve. Add and subtract so that one side of the equation is equal to zero. Factor the non-zero side just like normal. Set both inside and outside the parenthesis as equal to zero. Solve both equations for zero to get your final answer or answers. Plug your answers back in to ensure that they work.