Solving Quadratics Using Factorisation

Solving Simple Quadratics (only positive)

Solving quadratics using factorisation is usually the first method you learn for solving quadratic equations. Let’s start off by doing some simple quadratics that do not involve any negative numbers. In this fist example on the right, we have the equation x²+6x+8=0. We now have to factorise this equation (refer to factorisation to learn how to factorise double brackets). We can easily factorise x² by writing (x+_)(x+_). Now we only need to find the numbers that go into the brackets. To do this, we need to find the factors of 8 (1,8 & 2,4). We then have to check if they work with 6(x) by adding them together. 1+8=9 (wrong) and 2+4=6 (correct). Now we simply have to put 2 and 4 into the brackets and we have factorised the equation. Next comes the main step to solving the quadratic equation. We have to take the values in each pair of brackets and say that they are equal to 0 (so x+2=0 or x+4=0). Then we take the number in each linear equation to the other side of the equal sign to make x the subject (in this case x=-2 or x=-4). We now have the answers to the quadratic equation. To ensure that you have understood this concept, have a look at the examples underneath before moving on to solving negative quadratic using factorisation.
Example 1 We have solved the quadratic equation using factorisation. We first factorised x and then checked which factors of 32 make 12 when added together (4&8). We have then put these factors into the brackets before taking the values out of the brackets and making them equal to 0. This allowed us to find the two possible values of x.
Example 2 We have solved this quadratic equation by factorising it. We first factorised 2x which is (2x+_)(x+_). Then we found the factors of 8. Since we have 2x in the first bracket, we need to later on multiply one of the factors by 2. This is why 4 and 2 work. If we put 4 into the second pair of brackets, we will get 2 x 4 + 2 = 10. 10 is the number we need to make 10x. Then we just finish solving the equation as in the previous example just that we have to divide the values that were in the first bracket by 2 since we want to find the value of x and not 2x.

How to work with 2x², 3x², etc.

Just in case you have not understood example 2, we will look at how to deal with with 2x²,etc. in quadratic equations. Let’s have a look at the example on the right. The first step is quite simple. since we are dealing with 2x, we need to write the brackets as follows (2x+_)(x+_). If we expand these brackets, we will get 2x². However, since we have have 2x in the first bracket, we need to remember that the number in the second bracket has to be multiplied by 2. So when expanding the brackets, we would get 2x²+14x-6x-42. Although we write 7 in the bracket, when multiplying by 2x, it becomes 14x. Always consider this when solving quadratics. When you work with larger coefficients of x² such as 8x², you will have two different possibilities when putting the x values into the brackets: (8x+_)(x+_) (4x+_)(2x+_) You will have to see which one works in order to solve the equation.

Quadratics with One or Two Negatives

Sadly quadratic equations are rarely fully positive. This is why you have to be able to deal with negative quadratic equations as well.

Quadratics with Negative nx Value.

As you can see in the example on the right, we have -6x instead of +6x. Dealing with this is relatively simple. If 6x was positive, we would just look at the factors of 8, identify that 2 and 4 are the correct factors and put them into the brackets with + signs. In this case, we do pretty much the same just that we put - signs into both brackets. Like this, we get a negative nx value (-2-4=-6(x)) but a positive number at the end due to the fact that -2 x -4 = 8 (negative times negative = positive). Then you just finish solving the equation by taking the values out of the brackets and making x the subject.

Quadratics with Negative Number but Positive nx Value

In this example, we have a positive nx value but a negative number at the end. As a result of this, one bracket will have a minus sign and the other one a plus sign. This is because we have to multiply the factors to get the number at the end of the equation and the only way to get a negative number when multiplying is to times one positive number and one negative number together. We again look for factors making sure that we can create the nx value (in this case 1x). The factors 3 and 2 work if 3 is positive and 2 is negative because 3 x -2 = -6 and 3 - 2 = 1(x). So we get both the positive nx value and the negative number when expanding (x-2)(x+3).

Quadratics with Negative Number and nx Value

Now both the number and the nx value are negative. The method we use to solve this type of quadratic equation is quite similar to the one we used when we had a negative number and a positive nx value. The only real difference is that this time we are trying to make the nx value negative. We again start off by factorising x². Then we find the factors of 4. The factors we need to get -3x are 4 and 1. This time we want the negatives to win since -3x is negative. Consequently, we put a minus sign next to the bigger number and leave the smaller number positive. Now we just have to finish solving the equation as normal.


Here are four other quadratic equations one of each of the ones that we have learned about above.
Example 1 In this first example, we are dealing with a simple quadratic equation with only plus signs. We solved it by factorising x  and then finding the correct factors of 60 that give us 16x at the end. Once we factorised the equation, we were able to solve it.
Example 2 This equation has a negative nx value but a positive number at the end. As a result, bot brackets must contain a minus sign as we need to multiply negative by negative to get a positive number. After we found the correct factors of 16, we were able to solve the equation.
Example 3 In example 3, we have a quadratic equation with a negative number but a positive nx value. To solve it, we need to factorise 2x² and then find the correct factors of 12 before placing them in the brackets considering the 2x² and the plus and minus signs. The we can find the solution.
Example 4 In this last example, we have a quadratic in which both the nx value and the number at the end is negative. To solve it, we must find the correct pair of factors and place it in the brackets considering the fact that we have a 2x² equation and the plus and minus signs. Then we can deduct the solution.

At the beginning, the most difficult thing is to find the common factors. You will have to insert the factors and see if they work when you expand the

brackets. When you get more used to quadratic equations, you will not have to think a lot. You just know from looking and experience which factors

work. Make sure to consider the + and - signs or the coefficient of x² if applicable.

The Quadratic Formula

In some cases, you cannot solve a quadratic by factorising it. In these cases, you have to use the quadratic formula. Make sure to have a look at how to

apply it. Please share this page if you like it or found it helpful!

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Chapter 17.1:  Learning Outcomes Students will learn how to solve simple quadratics (only positive) using factorisation! Students will learn how to solve complex quadratics (involving one or more negatives) using factorisation!



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