Factoring Problem Solving

Put another way, the only way for us to get zero when we multiply two (or more) factors together is for one of the factors to have been zero.So, if we multiply two (or more) factors and get a zero result, then we know that at least one of the factors was itself equal to zero.We now include all the solutions we found in a single solution to the original problem: Equation Calculator - Solve By Factoring Solves an entered equation by factoring, showing step-by-step work.

We are left with three factors: 5x, (x 3), and (x - 3). " section, at least one of the three factors must be equal to zero.

Create three subproblems by setting each factor equal to zero.

This lesson covers many ways to solve quadratics, such as taking square roots, completing the square, and using the Quadratic Formula. (Before reaching the topic of solving quadratic equations, you should already know how to factor quadratic expressions.

If not, first review how to factor quadratics.) You've already factored quadratic expressions.

But what about equations where the variable carries an exponent, like x Solving the first subproblem, x - 2 = 0, gives x = 2. You should observe that as long as a does not equal 0, b must be equal to zero.

Solving the second subproblem, x - 3 = 0, gives x = 3. To state the observation more generally, "If ab = 0, then either a = 0 or b = 0." This is an important property of zero which we exploit when solving by factoring.The new thing here is that the quadratic expression is part of an equation, and you're told to solve for the values of the variable that make the equation true.Here's how it works: of those things that we multiplied must also have been equal to zero. And we have s squared minus 2s minus 35 is equal to 0. So when you do something by grouping, when you factor by grouping, you think about two numbers whose sum is going to be equal to negative 2. So, the fact that this number times that number is equal to zero tells us that either s plus 5 is equal to 0 or-- and maybe both of them-- s minus 7 is equal to 0. And so you have these two equations, and actually, we could say and/or.Now if this is the first time that you've seen this type of what's essentially a quadratic equation, you might be tempted to try to solve for s using traditional algebraic means, but the best way to solve this, especially when it's explicitly equal to 0, is to factor the left-hand side, and then think about the fact that those binomials that you factor into, that they have to be equal to 0. So you think about two numbers whose sum, a plus b, is equal to negative 2 and whose product is going to be equal to negative 35. So if the product is a negative number, one has to be positive, one has to be negative. So you have s plus 5 times this s right here, right? It could be or/and, either way, and both of them could be equal to 0. Well, we can just subtract 5 from both sides of this equation right there.The final step is to combine the two previous solutions, x = 2 and x = 3, into one solution for the original problem. When the example was factored into (x - 2)(x - 3) = 0, this property was applied to determine that either (x - 2) must equal zero, or (x - 3) must equal zero.Therefore, we were able to create two equations and determine two solutions from this observation.Several previous lessons explain the techniques used to factor expressions.This lesson focuses on an imporatant application of those techniques - solving equations.This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. Now, we call it factoring by grouping because we group it. And these first two terms, they have a common factor of s. And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. If I just have x plus a times x plus b, what is that equal to? We have 1 as a leading coefficient here, we have 1 as a leading coefficient here. We could have done that straight away and would've gotten to that right there. Now that we've factored it, we just have to think a little bit about what happens when you take the product of two numbers? If ever told you that I had two numbers, if I told you that I had the numbers a times b and that they equal to 0, what do we know about either a or b or both of them?