How to find the roots of a quadratic equation by factoring

Question

In pre-calculus, you can use the zero-product property to find the roots of a factored equation. After you factor a polynomial into its different pieces, you can set each piece equal to zero to solve for the roots with the zero-product property. The zero-product property says that if several factors are multiplying to give you zero, at least one of them has to be zero. Your job is to find all the values of x that make the polynomial equal to zero. This task is much easier if the polynomial is factored because you can set each factor equal to zero and solve for x.

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How do you find the roots using factoring?

Factoring x2 + 3x – 10 = 0 gives you (x + 5)(x – 2). Moving forward is easy because each factor is linear (first degree). The term x + 5 = 0 gives you one solution, x = –5, and x – 2 = 0 gives you the other solution, x = 2.

These solutions each become an x-intercept on the graph of the polynomial.

Sometimes after you’ve factored, one or both of the two factors can be factored again, in which case you should continue factoring. In other cases, they may be unfactorable. If one of these factors is a quadratic, you can find the roots only by using the quadratic formula. For example, 6x4 – 12x3 + 4x2 = 0 factors to 2x2(3x2 – 6x + 2) = 0. The first term, 2x2 = 0, is solvable using algebra, but the second factor, 3x2 – 6x + 2 = 0, is unfactorable and requires the quadratic formula.

In other cases, they may unfactorable, in which case you can solve them only by using the quadratic formula.

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Pre-Calculus ,

To solve an quadratic equation using factoring :

1 . Transform the equation using standard form in which one side is zero.

2 . Factor the non-zero side.

3 . Set each factor to zero (Remember: a product of factors is zero if and only if one or more of the factors is zero).

4 . Solve each resulting equation.

Example 1:

Solve the equation, x 2 − 3 x − 10 = 0

Factor the left side: ( x − 5 ) ( x + 2 ) = 0

Set each factor to zero: x − 5 = 0 or x + 2 = 0

Solve each equation: x = 5 or x = − 2

The solution set is { 5 , − 2 } .

Example 2:

Solve the equation, 2 x 2 + 5 x = 12

Set the right side to zero: 2 x 2 + 5 x − 12 = 0

Factor the left side: ( 2 x − 3 ) ( x + 4 ) = 0

Set each factor to zero: 2 x − 3 = 0 or x + 4 = 0

Solve each equation: x = 3 2 or x = − 4

The solution set is { 3 2 , − 4 } .

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Video transcript

We're asked to solve for s. And we have s squared minus 2s minus 35 is equal to 0. 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 let's just do that. So how can we factor this? We've seen it in several ways. I'll show you the standard we've been doing it, by grouping, and then there's a little bit of a shortcut when you have a 1 as a coefficient over here. 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 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. a times b is equal to negative 35. So if the product is a negative number, one has to be positive, one has to be negative. And so if you think about it, ones that are about two apart, you have 5 and negative 7, I think that'll work. 5 plus negative 7 is equal to negative 2. So to factor by grouping, you split this middle term. We can split this into a-- let me write it this way. We have s squared, and then this middle term right here, I'll do it in pink. This middle term right there I can write it as plus 5s minus 7s and then we have the minus 35. And then, of course, all of that is equal to 0. Now, we call it factoring by grouping because we group it. So we can group these first two terms. And these first two terms, they have a common factor of s. So let's factor that out. You have s times s plus 5. That's the same thing as s squared plus 5s. Now, in these second two terms right here, you have a common factor of negative 7, so let's factor that out. So you have negative 7 times s plus 5. And, of course, all of that is equal to 0. Now, we have two terms here, where both of them have s plus 5 as a factor. So we can factor that out. So let's do that. So you have s plus 5 times this s right here, right? S plus 5 times s will give you this term. And then you have minus that 7 right there. I undistributed the s plus 5. And then this is going to be equal to 0. 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? I mean, s plus 5 is a number. s minus 7 is another number. And we're saying that the product of those two numbers is equal to zero. 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? Well, at least one of them has to be equal to 0, or both of them have to be equal to 0. 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. I'll do that in just green. And so you have these two equations, and actually, we could say and/or. It could be or/and, either way, and both of them could be equal to 0. So let's see how we can solve for this. Well, we can just subtract 5 from both sides of this equation right there. And so you get, on the left-hand side, you have s is equal to negative 5. That is one solution to the equation, or you can add 7 to both sides of that equation, and you get s is equal to 7. So if s is equal to negative 5, or s is equal to 7, then we have satisfied this equation. We can even verify it. If you make s equal to negative 5, you have positive 25 plus 10, which is minus 35. That does equal zero. If you have 7, 49 minus 14 minus 35 does equal zero. So we've solved for s. Now, I mentioned there's an easier way to do it. 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. Let me just show you an example. If I just have x plus a times x plus b, what is that equal to? x times x is x squared, x times b is bx. a times x is plus ax. a times b is ab. So you get x squared plus-- these two can be added-- plus a plus bx plus ab. And that's the pattern that we have right here. We have 1 as a leading coefficient here, we have 1 as a leading coefficient here. So once we have our two numbers that add up to negative 2, that's our a plus b, and we have our product that gets to negative 35, then we can straight just factor it into the product of those two things. So it will be-- or the product of the binomials, where those will be the a's and the b's. So we figured it out. It's 5 and negative 7. 5 plus negative 7 is negative 2. 5 times negative 7 is negative 35. So we could have just straight factored at this point. 2, well, actually this was the case of s. So we could have factored it straight to the case of s plus 5 times s minus 7. We could have done that straight away and would've gotten to that right there. And, of course, that whole thing was equal to zero. So that would've been a little bit of a shortcut, but factoring by grouping is a completely appropriate way to do it as well.

How do you find the roots using factoring?

Since the roots of a function are the points at which y = 0, we can find the roots of y = ax² + bx + c = 0 by factoring ax² + bx + c = 0 and solving for x. We can also find the roots of y = ax² + bx + c = 0 using the quadratic formula, and we can find the number of roots using the discriminant.

find roots quadratic equation factoring