In this topic we combine factoring with solving an equation involving trinomials. Recall that solving means finding a value that makes the equation true. The focus is on equations that involve trinomials: \(ax^2+bx+c=0\).
Before we solve, we need to discuss the Zero Product Property. In words, if two numbers are multiplied together and give zero, then one of those two numbers had to be zero to start with. In math language:
$$\text{if } m\cdot n=0 \implies \text{either } m=0 \text{ or } n=0$$This is key to solving an equation that has trinomials in it.
Solve \(w^2+14w-32=0\).
| Factor the trinomial: | \(\implies\left(w+16\right)\left(w-2\right)=0\) |
| Apply Zero Product Property: | \(\implies w+16=0\text{ or }w-2=0\) |
| Get solutions: | \(\implies w=-16\text{ or }w=2\) |
Solve the equation: \(196z^2-49=0\).
| This problem is a difference of squares problem since \(196=\left(14\right)^2\) and \(49=(7)^2\) |
| \(\implies\left(14z+7\right)\left(14z-7\right)=0\) |
| \(\implies14z+7=0\text{ and }14z-7=0\) |
| \(\implies z=-\frac{14}{7}\text{ and }z=\frac{14}{7}\) |
Solve the equation: \(20x^2-29x-33=0\).
| Here is a difficult factoring problem, so use what method you need to get to the factored form: |
| \(\implies\left(5x+11\right)\left(4x-3\right)=0\) |
| \(\implies5x+11=0\text{ and }4x-3=0\) |
| \(\implies x=-\frac{11}{5}\text{ and }x=\frac{3}{4}\) |
All the equations shown so far have been equal to zero. Most equations will need some algebra work done in order to make them equal zero first.
Solve the equation: \(12t^2-50t+34=10t-38\).
| First, algebra is needed to get 0: subtract \(10t\) and add \(38\). | \(12t^2-50t \color{blue}{-10t}+72\color{blue}{+10}=0\) |
| \(\implies12t^2-60t+72=0\) | |
| Now factor: | \(\implies\left(6t-18\right)\left(2t-4\right)=0\) |
| Apply Zero Product Property: | \(\implies6t-18=0\text{ and }2t-4=0\) |
| \(\implies t=3\text{ and }t=2\) |
Solving relies heavily on getting the correct factored form so make sure you can factor completely. Below are some more helpful tips for success: