Chapter 13: Factorisation
Class 8 Maths Factorisation involves breaking expressions into simpler factors. Class 8 Maths Factorisation Notes explain methods such as common factor, regrouping, using identities, and dividing polynomials. Students learn to factorize trinomials and algebraic expressions using step-by-step techniques.
These notes also include important formulas like (a + b)² and (a – b)². With plenty of practice questions and solutions, students develop a strong base for algebra. They help enhance logical reasoning and speed in simplifying expressions for school exams.
Polynomial
A polynomial is a sum of one or more terms, where each term consists of a constant and one or more variables raised to some non-negative integer exponents. A polynomial with only one term is called a monomial, two terms, a binomial and three terms, a trinomial. Some examples of each type are listed below.
Types of Polynomials:
- Monomials: x, 1, 3x⁵y¹²
- Binomials: x + 3, 2x² + 5xy²
- Trinomials: x² + 5x + 29, 2x⁷ + 40y³z² + x²z⁵
Degree of a Polynomial
The degree of an individual term in a polynomial is the sum of the exponents of the variables in that term. For example, the term x²y³ has a degree of 5, since the exponents of the variables x and y (2 and 3 respectively) add up to 5.
Factoring
Factoring is the process by which we go about determining what we multiplied to get the given quantity. An important part of this process is being able to express a complicated polynomial into a product of simpler polynomials. This involves factorisation.
We do this all the time with numbers. For instance, here are a variety of ways to factor 12:
12 = (2)(6)
12 = (3)(4)
12 = (2)(2)(3)
12 = (-2)(-6)
12 = (-2)(2)(-3)
There are many more possible ways to factor 12, but these are representative of many of them.
Prime Numbers and Complete Factorization
A common method of factoring numbers is to completely factor the number into positive prime factors. A prime number is a number whose only positive factors are 1 and itself. For example 2, 3, 5, and 7 are all examples of prime numbers. Examples of numbers that aren't prime are 4, 6, and 12 to pick a few.
If we completely factor a number into positive prime factors there will only be one way of doing it. That is the reason for factoring things in this way. For our example above with 12 the complete factorization is:
12 = (2)(2)(3)
Examples of Complete Factorization
Example 1:
x² - 16 = (x + 4)(x - 4)
This is completely factored since neither of the two factors on the right can be further factored.
Example 2:
x⁴ - 16 = (x² + 4)(x² - 4)
This is NOT completely factored because the second factor can be further factored. Note that the first factor is completely factored however. Here is the complete factorization of this polynomial:
x⁴ - 16 = (x² + 4)(x + 2)(x - 2)
The purpose of this section is to familiarize ourselves with many of the techniques for factoring polynomials.
Factoring Binomials
The simplest method of factoring binomials is to remove factors that are common to different terms. This method does not work in all situations. The formulas listed below can be used to factor binomials that fit the specific situation.
Important Formulas for Binomials
- Difference of Squares: x² - y² = (x - y)(x + y)
- Sum of Cubes: x³ + y³ = (x + y)(x² - xy + y²)
- Difference of Cubes: x³ - y³ = (x - y)(x² + xy + y²)
Examples - Factoring Binomials
Example (i): Factor 2x³ + 10x² - 4x
Solution:
2x³ + 10x² - 4x = 2x(x² + 5x - 2)
Example (ii): Factor x³y³ + 2x²y + 2xy² + xy
Solution:
x³y³ + 2x²y + 2xy² + xy = xy(x²y² + 2x + 2y + 1)
Example (iv): Factor a³ + 27b⁶c
Solution:
a³ + 27b⁶c = a³ + (3b²)³c
Using sum of cubes formula:
= (a + 3b²c)(a² - 3ab²c + 9b⁴c²)
Factoring Trinomials
Quadratic trinomials have the form ax² + bx + c. When factoring these trinomials, we want to write them as a product of two binomials. The two types of quadratic trinomials are listed below.
Monic Quadratic Trinomials
Monic quadratic trinomials are expressions where the leading coefficient (a) is equal to 1. For example, x² + 7x + 12 is a monic quadratic trinomial. These trinomials are the simplest to factor. For the general monic quadratic trinomial, x² + bx + c, we must find the roots of the polynomial, x₁ and x₂, such that x² + bx + c = (x - x₁)(x - x₂). By expanding the factorization, we see that x² + bx + c = x² - (x₁ + x₂)x + (x₁x₂). From this expansion, we get the formulas b = -(x₁ + x₂) and c = x₁x₂. We can use these formulas to find the roots of the polynomial, if it can be factored.
The simplest way to factor monic quadratic trinomials is to look for integer roots. If we cannot find integer roots however, it does not mean that the trinomial can't be factored. We must use the quadratic formula, which is given in the quadratic equations tutorial, to find the non-integer roots of the quadratic trinomial. If no roots exist, however, then the trinomial cannot be factored.
Important Things to Remember When Factoring Trinomials
Using a multiplication problem consisting of two binomials, we will show some important things to remember when factoring trinomials, which is the reverse of multiplying two binomials.
Example: (x - 6)(x + 3) = x² - 6x + 3x - 18 = x² - 3x - 18
Keeping these important things in mind, you can factor trinomials:
- The first term of the trinomial is the product of the first terms of the binomials.
- The last term of the trinomial is the product of the last terms of the binomials.
- The coefficient of the middle term of the trinomial is the sum of the last terms of the binomials.
- If all the signs in the trinomial are positive, all signs in both binomials are positive.
Solved Examples - Monic Trinomials
1. Factor: x² - 14x - 15
Solution:
First, write down two sets of parentheses to indicate the product.
( )( )
Since the first term in the trinomial is x², the product of the first terms of the binomials, you enter x as the first term of each binomial.
(x )(x )
The product of the last terms of the binomials must equal -15, and their sum must equal -14, and one of the binomials' terms has to be negative.
Four different pairs of factors have a product that equals -15:
(3)(-5) = -15 (-15)(1) = -15
(-3)(5) = -15 (15)(-1) = -15
However, only one of those pairs has a sum of -14.
(-15) + (1) = -14
Therefore, the second terms in the binomial are -15 and 1 because these are the only two factors whose product is -15 (the last term of the trinomial) and whose sum is -14 (the coefficient of the middle term in the trinomial).
Answer: (x - 15)(x + 1)
2. Factor x² + 6x + 8
Solution:
To factor this monic quadratic equation we must find roots x₁ and x₂, such that -(x₁ + x₂) = 6 and x₁x₂ = 8. So we must find two numbers whose sum is -6 and whose product is 8. Clearly, -2 and -4 add to give -6 and multiply to give 8. We can now factor the expression.
x² + 6x + 8 = (x - (-2))(x - (-4))
x² + 6x + 8 = (x + 2)(x + 4)
3. Factor y² + 4y - 21
Solution:
To factor this monic quadratic equation, we must find roots y₁ and y₂, such that -(y₁ + y₂) = 4 and y₁y₂ = -21. So we must find the numbers whose sum is -4 and whose product is -21. Clearly, -7 and 3 add to give -4 and multiply to give -21. We can now factor the expression.
y² + 4y - 21 = (y - (-7))(y - (3))
y² + 4y - 21 = (y + 7)(y - 3)
Nonmonic Quadratic Trinomials
Nonmonic quadratic trinomials are expressions where the leading coefficient (a) is not equal to 1. For example 6x² - 11x - 7 is a nonmonic quadratic trinomial. To factor these trinomials, we must again find roots x₁ and x₂ such that ax² + bx + c = a(x - x₁)(x - x₂).
Method: The simplest way to factor a nonmonic trinomial is to find two integers m and n, such that m + n = b and m × n = ac. After finding the two integers, substitute the bx term for the sum of terms mx + nx. Then remove common factors. If no integers can be found, it does not necessarily mean that the polynomial cannot be factored. We must find the non-integer roots of the polynomial using the quadratic formula. If there are no roots of the polynomial, it cannot be factored.
4. Factor 6x² - 11x - 7
Solution:
We need to find integers m, n such that m + n = -11 and m × n = (6)(-7) = -42. By writing out all the factors of -42, we will find that -14 and 3 multiply to give -42 and add to give -11.
6x² - 11x - 7 = 6x² - 14x + 3x - 7
= 2x(3x - 7) + 3x - 7
= 2x(3x - 7) + 1(3x - 7)
= (3x - 7)(2x + 1)
So 6x² - 11x - 7 = (3x - 7)(2x + 1)
5. Factor 4x² + 13x - 5
Solution:
We need to find integers m, n such that m + n = 13 and m × n = (4)(-5) = -20.
By writing out all the factors of -20, we find that there are no integers that add to give 13 and multiply to give -20.
| 1 | 2 | 4 | 5 | 10 | 20 |
| -20 | -10 | -5 | -4 | -2 | -1 |
| -19 | -8 | -1 | 1 | 8 | 19 |
So 4x² + 13x - 5 cannot be simply factored. In order to factor the polynomial, we would need to find the roots using the quadratic formula.
Greatest Common Factor (GCF)
The first method for factoring polynomials will be factoring out the greatest common factor. When factoring in general this will also be the first thing that we should try as it will often simplify the problem.
To use this method all that we do is look at all the terms and determine if there is a factor that is in common to all the terms. If there is, we will factor it out of the polynomial. Also note that in this case we are really only using the distributive law in reverse. Remember that the distributive law states that:
a(b + c) = ab + ac
In factoring out the greatest common factor we do this in reverse. We notice that each term has an a in it and so we "factor" it out using the distributive law in reverse as follows:
ab + ac = a(b + c)
Examples - Greatest Common Factor
1. Factor out the greatest common factor from each of the following polynomials.
(a) 8x⁴ - 4x³ + 10x²
Solution:
First we will notice that we can factor a 2 out of every term. Also note that we can factor an x² out of every term. Here then is the factoring for this problem.
8x⁴ - 4x³ + 10x² = 2x²(4x² - 2x + 5)
Note that we can always check our factoring by multiplying the terms back out to make sure we get the original polynomial.
(b) x³y² + 3x⁴y + 5x⁵y³
Solution:
In this case we have both x's and y's in the terms but that doesn't change how the process works. Each term contains an x³ and a y so we can factor both of those out. Doing this gives:
x³y² + 3x⁴y + 5x⁵y³ = x³y(y + 3x + 5x²y²)
(c) 3x⁶ - 9x² + 3x
Solution:
In this case we can factor a 3x out of every term. Here is the work for this one.
3x⁶ - 9x² + 3x = 3x(x⁵ - 3x + 1)
Notice the "+1" where the 3x originally was in the final term, since the final term was the term we factored out we needed to remind ourselves that there was a term there originally. To do this we need the "+1" and notice that it is "+1" instead of "-1" because the term was originally a positive term. If it had been a negative term originally we would have had to use "-1".
One of the more common mistakes with these types of factoring problems is to forget this "1". Remember that we can always check by multiplying the two back out to make sure we get the original. To check that the "+1" is required, let's drop it and then multiply out to see what we get.
3x(x⁵ - 3x) = 3x⁶ - 9x² ≠ 3x⁶ - 9x² + 3x
So, without the "+1" we don't get the original polynomial! Be careful with this. It is easy to get in a hurry and forget to add a "+1" or "-1" as required when factoring out a complete term.
(d) 9x²(2x + 7) - 12x(2x + 7)
Solution:
This one looks a little odd in comparison to the others. However, it works the same way. There is a 3x in each term and there is also a 2x + 7 in each term and so that can also be factored out. Doing the factoring for this problem gives:
9x²(2x + 7) - 12x(2x + 7) = 3x(2x + 7)(3x - 4)
Factoring by Grouping
This is a method that isn't used all that often, but when it can be used it can be somewhat useful. This method is best illustrated with an example or two.
1. Factor by grouping each of the following.
(a) 3x² - 2x + 12x - 8
Solution:
In this case we group the first two terms and the final two terms as shown here:
(3x² - 2x) + (12x - 8)
Now, notice that we can factor an x out of the first grouping and a 4 out of the second grouping. Doing this gives:
3x² - 2x + 12x - 8 = x(3x - 2) + 4(3x - 2)
We can now see that we can factor out a common factor of 3x - 2 so let's do that to the final factored form.
3x² - 2x + 12x - 8 = (3x - 2)(x + 4)
And we're done. That's all that there is to factoring by grouping. Note again that this will not always work and sometimes the only way to know if it will work or not is to try it and see what you get.
(b) x⁵ + x - 2x⁴ - 2
Solution:
In this case we will do the same initial step, but this time notice that both of the final two terms are negative so we'll factor out a "-" as well when we group them. Doing this gives:
(x⁵ + x) - (2x⁴ + 2)
We can always distribute the "-" back through the parenthesis to make sure we get the original polynomial.
At this point we can see that we can factor an x out of the first term and a 2 out of the second term. This gives:
x⁵ + x - 2x⁴ - 2 = x(x⁴ + 1) - 2(x⁴ + 1)
We now have a common factor that we can factor out to complete the problem.
x⁵ + x - 2x⁴ - 2 = (x⁴ + 1)(x - 2)
(c) x⁵ - 3x³ - 2x² + 6
Solution:
This one also has a "-" in front of the third term as we saw in the last part. However, this time the fourth term has a "+" in front of it unlike the last part. We will still factor a "-" out when we group however to make sure that we don't lose track of it. When we factor the "-" out notice that we needed to change the "+" on the fourth term to a "-". Again, you can always check that this was done correctly by multiplying the "-" back through the parenthesis.
(x⁵ - 3x³) - (2x² - 6)
Now that we've done a couple of these we won't put the remaining details in and we'll go straight to the final factoring.
x⁵ - 3x³ - 2x² + 6 = x³(x² - 3) - 2(x² - 3) = (x² - 3)(x³ - 2)
Factoring by grouping can be nice, but it doesn't work all that often. Notice that as we saw in the last two parts of this example if there is a "-" in front of the third term we will often also factor that out of the third and fourth terms when we group them.
Factoring Quadratic Polynomials
First, let's note that quadratic is another term for second degree polynomial. So we know that the largest exponent in a quadratic polynomial will be a 2. In these problems we will be attempting to factor quadratic polynomials into two first degree (henceforth linear) polynomials. Until you become good at these, we usually end up doing these by trial and error although there are a couple of processes that can make them somewhat easier.
Detailed Examples - Factoring Quadratic Polynomials
1. Factor each of the following polynomials.
(a) x² + 2x - 15
Solution:
Okay since the first term is x² we know that the factoring must take the form.
x² + 2x - 15 = (x + _)(x + _)
We know that it will take this form because when we multiply the two linear terms the first term must be x² and the only way to get that to show up is to multiply x by x. Therefore, the first term in each factor must be an x. To finish this we just need to determine the two numbers that need to go in the blank spots.
We can narrow down the possibilities considerably. Upon multiplying the two factors out these two numbers will need to multiply out to get -15. In other words these two numbers must be factors of -15. Here are all the possible ways to factor -15 using only integers.
(-1)(15) (1)(-15) (-3)(5) (3)(-5)
Now, we can just plug these in one after another and multiply out until we get the correct pair. However, there is another trick that we can use here to help us out. The correct pair of numbers must add to get the coefficient of the x term. So, in this case the third pair of factors will add to "+2" and so that is the pair we are after.
Here is the factored form of the polynomial.
x² + 2x - 15 = (x - 3)(x + 5)
Again, we can always check that we got the correct answer by doing a quick multiplication.
Note that the method we used here will only work if the coefficient of the x² term is one. If it is anything else this won't work and we really will be back to trial and error to get the correct factoring form.
(b) x² - 10x + 24
Solution:
Let's write down the initial form again:
x² - 10x + 24 = (x + _)(x + _)
Now, we need two numbers that multiply to get 24 and add to get -10. It looks like -6 and -4 will do the trick and so the factored form of this polynomial is:
x² - 10x + 24 = (x - 4)(x - 6)
(c) x² + 6x + 9
Solution:
Again, let's start with the initial form:
x² + 6x + 9 = (x + _)(x + _)
This time we need two numbers that multiply to get 9 and add to get 6. In this case 3 and 3 will be the correct pair of numbers. Don't forget that the two numbers can be the same number on occasion as they are here.
Here is the factored form for this polynomial.
x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)²
Note as well that we further simplified the factoring to acknowledge that it is a perfect square. You should always do this when it happens.
(d) x² + 5x + 1
Solution:
Once again, here is the initial form:
x² + 5x + 1 = (x + _)(x + _)
Okay, this time we need two numbers that multiply to get 1 and add to get 5. There aren't two integers that will do this and so this quadratic doesn't factor.
This will happen on occasion so don't get excited about it when it does.
(e) 3x² + 2x - 8
Solution:
Okay, we no longer have a coefficient of 1 on the x² term. However we can still make a guess as to the initial form of the factoring. Since the coefficient of the x² term is a 3 and there are only two positive factors of 3 there is really only one possibility for the initial form of the factoring.
3x² + 2x - 8 = (3x + _)(x + _)
Since the only way to get a 3x² is to multiply a 3x and an x these must be the first two terms. However, finding the numbers for the two blanks will not be as easy as the previous examples. We will need to start off with all the factors of -8.
(-1)(8) (1)(-8) (-2)(4) (2)(-4)
At this point the only option is to pick a pair plug them in and see what happens when we multiply the terms out. Let's start with the fourth pair. Let's plug the numbers in and see what we get.
(3x + 2)(x - 4) = 3x² - 10x - 8
Well the first and last terms are correct, but then they should be since we've picked numbers to make sure those work out correctly. However, since the middle term isn't correct this isn't the correct factoring of the polynomial.
That doesn't mean that we guessed wrong however. With the previous parts of this example it didn't matter which blank got which number. This time it does. Let's flip the order and see what we get.
(3x - 4)(x + 2) = 3x² + 2x - 8 ✓
So, we got it. We did guess correctly the first time we just put them into the wrong spot.
So, in these problems don't forget to check both places for each pair to see if either will work.
(f) 5x² - 17x + 6
Solution:
Again the coefficient of the x² term has only two positive factors so we've only got one possible initial form.
5x² - 17x + 6 = (5x + _)(x + _)
Next we need all the factors of 6. Here they are.
(1)(6) (-1)(-6) (2)(3) (-2)(-3)
Don't forget the negative factors. They are often the ones that we want. In fact, upon noticing that the coefficient of the x is negative we can be assured that we will need one of the two pairs of negative factors since that will be the only way we will get negative coefficient there. With some trial and error we can get that the factoring of this polynomial is:
5x² - 17x + 6 = (5x - 2)(x - 3)
(g) 4x² + 10x - 6
Solution:
In this final step we've got a harder problem here. The coefficient of the x² term now has more than one pair of positive factors. This means that the initial form must be one of the following possibilities.
4x² + 10x - 6 = (4x + _)(x + _)
4x² + 10x - 6 = (2x + _)(2x + _)
To fill in the blanks we will need all the factors of -6. Here they are:
(-1)(6) (1)(-6) (-2)(3) (2)(-3)
With some trial and error we can find that the correct factoring of this polynomial is:
4x² + 10x - 6 = (2x - 1)(2x + 6)
Note as well that in the trial and error phase we need to make sure and plug each pair into both possible forms and in both possible orderings to correctly determine if it is the correct pair of factors or not.
We can actually go one more step here and factor a 2 out of the second term if we'd like to. This gives:
4x² + 10x - 6 = 2(2x - 1)(x + 3)
This is important because we could also have factored this as:
4x² + 10x - 6 = (4x - 2)(x + 3)
which, on the surface, appears to be different from the first form given above. However, in this case we can factor a 2 out of the first term to get:
4x² + 10x - 6 = 2(2x - 1)(x + 3)
This is exactly what we got the first time and so we really do have the same factored form of this polynomial.
Special Forms
There are some nice special forms of some polynomials that can make factoring easier for us on occasion. Here are the special forms.
Perfect Square Trinomial (Positive):
a² + 2ab + b² = (a + b)²
Perfect Square Trinomial (Negative):
a² - 2ab + b² = (a - b)²
Difference of Squares:
a² - b² = (a + b)(a - b)
Sum of Cubes:
a³ + b³ = (a + b)(a² - ab + b²)
Difference of Cubes:
a³ - b³ = (a - b)(a² + ab + b²)
Examples - Special Forms
1. Factor each of the following.
(a) x² - 20x + 100
Solution:
In this case we've got three terms and it's a quadratic polynomial. Notice as well that the constant is a perfect square and its square root is 10. Notice as well that 2(10) = 20 and this is the coefficient of the x term. So, it looks like we've got the second special form above. The correct factoring of this polynomial is:
x² - 20x + 100 = (x - 10)²
To be honest, it might have been easier to just use the general process for factoring quadratic polynomials in this case rather than checking that it was one of the special forms, but we did need to see one of them worked.
(b) 25x² - 9
Solution:
In this case all that we need to notice is that we've got a difference of perfect squares:
25x² - 9 = (5x)² - (3)²
So, this must be the third special form above. Here is the correct factoring for this polynomial.
25x² - 9 = (5x + 3)(5x - 3)
(c) 8x³ + 1
Solution:
This problem is the sum of two perfect cubes:
8x³ + 1 = (2x)³ + (1)³
and so we know that it is the fourth special form from above. Here is the factoring for this polynomial.
8x³ + 1 = (2x + 1)(4x² - 2x + 1)
a² + b² ≠ (a + b)²
This just simply isn't true for the vast majority of sums of squares, so be careful not to make this very common mistake. There are rare cases where this can be done, but none of those special cases will be seen here.
Factoring Polynomials with Degree Greater Than 2
There is no one method for doing these in general. However, there are some that we can do so let's take a look at a couple of examples.
1. Factor each of the following.
(a) 3x⁴ - 3x³ - 36x²
Solution:
In this case let's notice that we can factor out a common factor of 3x² from all the terms so let's do that first.
3x⁴ - 3x³ - 36x² = 3x²(x² - x - 12)
What is left is a quadratic that we can use the techniques from above to factor. Doing this gives us:
3x⁴ - 3x³ - 36x² = 3x²(x - 4)(x + 3)
Don't forget that the FIRST step to factoring should always be to factor out the greatest common factor. This can only help the process.
(b) x⁴ - 25
Solution:
There is no greatest common factor here. However, notice that this is the difference of two perfect squares.
x⁴ - 25 = (x²)² - (5)²
So, we can use the third special form from above.
x⁴ - 25 = (x² + 5)(x² - 5)
Neither of these can be further factored and so we are done. Note however, that often we will need to do some further factoring at this stage.
(c) x⁴ + x² - 20
Solution:
Let's start this off by working a factoring a different polynomial.
u² + u - 20 = (u - 4)(u + 5)
We used a different variable here since we'd already used x's for the original polynomial.
So, why did we work this? Well notice that if we let u = x² then u² = (x²)² = x⁴. We can then rewrite the original polynomial in terms of u's as follows:
x⁴ + x² - 20 = u² + u - 20
and we know how to factor this. So factor the polynomial in u's then back substitute using the fact that we know u = x².
x⁴ + x² - 20 = u² + u - 20
= (u - 4)(u + 5)
= (x² - 4)(x² + 5)
Finally, notice that the first term will also factor since it is the difference of two perfect squares. The correct factoring of this polynomial is then:
x⁴ + x² - 20 = (x - 2)(x + 2)(x² + 5)
Note that this converting to u first can be useful on occasion, however once you get used to these this is usually done in our heads.
Division of A polynomial by Another Polynomial
You are already familiar with the division of a number by another number. You are also familiar with division algorithm which is:
Dividend = quotient × divisor + remainder
The process of division of a polynomial by another polynomial is also same as we use in number system. The process of division may be divided in three cases in polynomials.
Three Cases of Polynomial Division
- Division of a monomial by another monomial
- Division of a polynomial by monomial
- Division of a polynomial by another polynomial
Two Methods for Division
We have two methods for division of a polynomial by another polynomial:
(a) Factor method: Factor method generally used when remainder is zero.
(b) Long division: This method generally used when remainder is not zero. This method can also used when remainder is zero. We shall discuss these three cases one by one.