We've learned about the greatest common factor and how to factor by grouping. Let's see if we can use that knowledge to solve this equation. Let's start by factoring the left side of the equation. Write your two factors in this box.
When we factor a quadratic trinomial, we want to find factors of a times c that sum to b. So in this case, we want to find factors of 1 times 6, or just 6, that sum to 5. I know 3 and 2 are factors of 6, that add to 5. So I can use those to rewrite my middle term. Then we use factoring by grouping to get x plus 2 times x plus 3. Great work if you got that one right. You might have noticed a pattern when factoring trinomials that have an a value of 1. The two factors here always appear in this parentheses. But keep in mind, this is only if the a value is 1.
We changed our equation to look like this now. Let's think about this new equation. We know x really represents some number so I have one number here added to 2 and I have another number here added to 3. We really have a number times a number equals 0. Which of these statements must be true about the two numbers? Choose the best choice.
This last choice is correct. Fantastic reasoning if you figured that one out. As an example, the first number could be 4 and the second one could be 0. We know 4 times 0 equals 0. And let's say the second number is 5. 0 times 5 equals 0. Now, I don't actually know that one of the numbers is 4 or the other is 5, but one of them must equal 0, since our product is 0.
When we multiply two number together and get zero, then we know that one of the numbers, a or b, must equal zero. This is called the Zero-Product Property. Let's use that knowledge and try and answer this question. What would the product be of all of these binomials? And if all these variables are confusing to you, just remember that all of these are numbers. Write your answer here.
Well the answer would be zero. This one was tough. So don't worry if you didn't get it right on your first try. We can see that one of the factors is x minus x. Well, we know x minus x would equal zero, so I have a bunch of other numbers, times zero, times a bunch of other numbers. Anything times zero is just zero. I think it's pretty amazing how one number can have such a unique property in math. This one zero makes this whole entire multiplication zero.
Let's get back to trying to solve our quadratic equation. We know either this first number can equal 0 or the second number could equal 0. So, we can set each factor equal to 0. We solve each equation and we know x could equal negative 2 or x could equal negative 3.
Just like linear equations, we can also Check a Quadratic Equation. After we solve for our two answers, negative 2 and negative 3, we can plug them back into our original equation to make sure we're correct. I want you to check our two solutions by plugging in the appropriate numbers into the parenthesis.
For the first one, we plug in negative 2 for x. The quantity negative 2 squared is 4. And 5 times negative 2 is negative 10. These numbers sum to negative 6. And these numbers sum to 0. For this check, we have the quantity negative 3 squared. Plus 5 times negative 3 plus 6 equals 0. The square of negative 3 is 9, and 5 times negative 3 is negative 15. These numbers sum to negative 6 and these numbers sum to 0.
Whenever we write the solution to a quadratic equation, we write it as a set. X could equal negative 3, or x could equal negative 2. Use curly braces on the outside, and commas between the elements. You'll also notice that I listed the negative 3 first. By convention, we list the numbers from least to greatest.
For the first practice, try finding the solution to this equation. Keep in mind that not all quadratic equations can be solved using factoring. This one can.
For this quadratic, we want to find factors of 1 times negative 4 that sum to 3. Negative 1 and 4, multiply, to give us negative 4 and sum to give us 3. Since the coefficient on the square term is just 1, I can use these factors for my factored form. Remember that this would not be the case if this was not a 1. We'd have to use factoring by grouping. Then, we use our 0 product property to set each factor equal to 0. Adding one to both sides, we get p is equal to 1. And subtracting four from both sides, we could get p is equal to negative 4. Great work if you got this solution set.
Just as we did when factoring trinomials, we want to factor out any greatest common factor before solving an equation. After we factor, we want to set each factor equal to zero, including our greatest common factor. For example, if we wanted to solve this equation, we'd want to remove a greatest common factor first. Do you see it? And it turns out that this equation will have three solutions. And you might be wondering why. Well, the highest power here is a three, and this highest power on the variable tells us how many solutions we'll have for an equation. So, since we have z cubed or z to the third, we'll have three solutions. So, what would be the answers here? What could z equal?
You might have done some mental math or you might have shown all the work. Either way, the final answer is negative 1, 0, and 1. First, I factor 6z out from both terms. This is the greatest common factor. Next, we factor this difference of two squares. You have a squared minus b squared. So I can write z plus 1 and Z minus 1. And we know the product of these factors equals 0. We set each of the factors equal to 0 and then we solve for z. So z could equal 0, negative 1, or positive 1.
So, we saw that we could get three solutions when we have a third power for a variable. In this case the three solutions were all different. But sometimes we can get the same solution more than once. For example, in this equation, one of the solutions will be repeated. Do you think you can find what they are?
Sometimes we'll need to change the equation before attempting to factor. What do you think would be the best step to perform next if we wanted to factor this equation? Choose one of these steps.
Well it turns out it would be best to subtract 30a from both sides. I know there's a lot of different options here, so lets see why. We've seen factoring quadratic equations in this form. The variable was x and the numbers were a, b, and c. In this case, my variable isn't x. It's a. And my numbers are 2, 30, and this form. 30a is not like with a squared or 112. So, I'm going to list it here, so that way it falls in the middle of the 2 terms. I'll get 0 on the right, and then 2a squared, minus 30a, plus 112 on the left. Now this is something I'm familiar with and we can factor it.
So you try and finish solving this problem, what would be the solutions to the equation? Write your solution as I said in this box. I'll admit, this one is a bit tricky. So try again if at first you don't succeed.
correct. When we start to factor a quadratic Equation we should look for the GCF first. All the coefficients are even, so we can remove a 2. If we factor a 2 ,we're left with a squared minus 15a plus 56. Now we want to find factors of 1 times 56. At sum to negative 15. This top number's positive, and this bottom number's negative. This must mean my two factors have to be negative. Negative 7 times negative 8 is positive 56. And negative 7 plus negative 8 is negative 15. Since the coefficient is 1 I can use these two factors in my factors. Next we set each factor equal to 0. We know 2 does not equal 0. So 2 is not a solution. When we solve these two equations we get a is equal to 7 and a is equal to 8.
Try solving this equation. As a hint, you'll want to distribute the p and then rearrange your equation. Write your answer here.
p could equal negative 2, or 1 4th, excellent effort if you got that one correct. First we distribute the p then we subtract 2 from both sides. The negative 2 does not have any liker terms. So we have 4p squared plus 7p minus 2 equals 0. We look for our greatest common factor, and the only common factor between these three is 1. So we can't factor anything out. Now I want to find factors of negative 8 that add to 7. The two factors are 8 and negative 1, they multiply to give us negative 8 and they sum to give us 7. These two factors allows to rewrite our middle term and then we can use factoring by grouping. We continue factoring our equation by factoring a p by the first group of terms, and a 2 from the last group of terms. Next we said each of these factors equal to 0. P could equal 2 or p could equal 1 4th. These are our solutions.
I know these are some tough problems, so keep up your effort. Try this one out. What could m equal in this case? Write your answer here.
Negative 2 and 5 are correct. This equation doesn't need to be rearranged. We just want to find factors of negative 10 at sum to negative 3. Negative 5 and 2 are the factor pair we need. We write our factors, set each one equal to zero and solve for m. So, m could equal 5 or m could equal negative 2. And remember, we can always check our answers. We can plug each answer into the original equation to check. I get true statements in the end, so yes, 5 and negative 2 are correct.
How about this equation? What could be the values of a?
Negative 1 and 3 halves, or 1 and a half, is correct. Excellent algebra skills for getting that one right. When we look at the quadratic equation, we want to look for our greatest common factor first. Each term shows a factor of 5. So I can factor that out. We can factor the 5 out, leaving us with 2a squared, minus a, minus 3. Remember to keep writing this as an equation. All of these factors multiplied together equal 0. Next I want to find factors of negative 6 that sum to negative 1. I can use negative 3 and positive 2. We remove an a from the first group of terms and a positive 1 from the second group of terms. There's no other greatest common factor for these two terms except 1. I notice that I have We factor 2a minus 3 from this first term and the second term to leave us with a product of factors. Next we set each factor equal to 0, and then we solve for the values of a. We know 5 doesn't equal 0, so this is not a solution. A could equal negative 1, or a could equal 3 halves. And this would be our solution set.
For our challenge question I want you to try and solve this equation. What's the solution set?