Quadratic expressions are fundamental in mathematics and appear frequently in various applications, from physics to finance. Use this resource to learn what quadratics are and how to factorise them. The ability to do this is crucial for solving quadratic equations and simplifying complex algebraic problems.
Video tutorial – factorisation: quadratics
Watch this video to learn what quadratics are, review expansion, and understand how to factorise quadratics.
Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on factorising quadratics. The general form of a quadratic expression is given by A X squared plus B X plus C. Here A, B and C are constants and X is the unknown. For instance here’s an expression, two X squared minus three X minus four, A is plus two, B is minus three and C is minus four. Note carefully the sign for each of the numbers.
Let’s go back a step first and look at revision of expansion, in other words removing brackets. Here’s an expression, X plus two X plus three and when you expand that you multiply each term in the first bracket by the terms in the second bracket to get an answer of X squared plus five X plus six. We’ve done this in a previous movie.
Now factorisation is the reverse of expansion, in other words X square plus five X plus six can be expressed to the product of two factors, X plus two and X plus three. In other words as you go from left to right in this equation we’re factorising, as we go from right to left we’re expanding. A few important things to notice about factorisation; first of all multiplying the first terms in each bracket gives the X squared in the expression, in other words X times X is X squared. Multiplying the last terms in each bracket gives the constant term plus six in the expression, in other words plus two times plus three is plus six. Finally the co-efficient of the X terms, in other words the middle term, is the sum of the last term in each bracket, in other words here plus two and plus three is plus five. Now we’ll be using these steps to work out some of these examples that follow.
So in general for the expression A squared plus B X plus C where A is equal to one multiply two numbers to get C, we add the same two numbers to get B, in other words we insert one number into each of the brackets and secondly the order is not important, in other words the second brackets could be the first and vice versa, it doesn’t matter which way around you place the brackets.
So here’s our first example, X squared plus nine X plus 14. So straight away we can place two brackets, as I’ve done, to the right of the equal signs. Using the rules we need two numbers that when multiplied together gives us the last term, 14, and those same two numbers that, when added, gives us the co-efficient of X, the middle term, which is nine. So we can look at 14 and one, seven and two and so on until we get those two numbers that when multiplied gives us 14 but also added gives us nine and those two numbers are seven and two, seven times two is 14, seven plus two is nine. Notice if I use 14 and one multiplying them is fine but adding them will not give us nine.
Let’s look at another one; Y squared minus seven Y plus 12. Here we have a negative sign in the middle term which makes it slightly more difficult but we still use the same idea as in the previous example, in other words two numbers that multiply gives us 12, the same two numbers that when added gives us minus seven, so we can look at 12 and one, six and two, four and three. Notice when we look at those different combinations, those factors of 12, there’s only one that will satisfy the two conditions and that is minus three and minus four. Notice minus three times minus four is plus 12, minus three plus minus four is minus seven, so our answer is Y minus three Y minus four.
Here’s another one. Notice the terms are separated by negative signs, P squared minus five P minus 14. Again, using the rules, multiplying the two you must get minus 14, adding the two numbers we must get minus five. Here it’s not so difficult because we only have 14 and one, that gives us 14, but 14 and one there’s no way in which, when adding those two numbers, whether they be negative or positive, will give us minus five, so the solution to this is P minus seven and P plus two. Be extra careful with the negative signs here, notice that minus seven times plus two gives us minus 14 and minus seven plus two is minus five.
And finally here’s another one; A squared plus six A minus seven. We need two numbers that multiply to give minus seven and two numbers that add to give plus six. So we don’t have much choice here, the only factors we have of seven are seven and one, so they must be seven and one, but the difficult problem here is to work out where the negative signs go and this usually comes just with trial and error, in other words A plus seven A minus one, seven times minus one is minus seven, plus seven minus one is plus six. If you tried minus seven and plus one you would notice you would not get plus six A, so try the different methods to make sure you’re correct.
Now try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.
Quadratic expressions
Image: Pixabay
Sometimes, expansion results in a second-degree algebraic expression, or quadratic expression. This is when the highest power of an algebraic term is \(2\). The term "quadratic" comes from the Latin quad which means "to make square". Making a number a square is the same as taking it to the power of \(2\).
Some examples of quadratic expressions are:
\(x^{2}+5x+6\)
\(4x^{2}+8\)
\(2x^{2}-3x\)
\(7x^{2}-x+15\)
Compare this to a linear expression, where an algebraic term has a term with the highest power of \(1\).
Linear expressions have the form \(ax+by+c\). Quadratic expressions take the general form:
\[ax^{2}+bx+c\]
where \(a\), \(b\) and \(c\) are constants and \(a \neq 0\)
To start with, we will work with expressions that have \(a=1\) so the expression becomes \(x^{2}+bx+c\).
Example – expanding algebraic expressions
Expand \((x+2)(x+3)\).
To expand the expression, multiply each term in the first bracket by each term in the second bracket.
Just like factorising simpler algebraic expressions, factorising quadratics involves expressing them as products of their factors. This is the reverse of expansion.
The basic rule is:
\[x^{2}+bx+c = (x+m)(x+n)\]
where \(m\times n=c\) and \(m+n=b\)
\(x+m\) and \(m+n\) are the factors. The order in which they are written does not matter. That is:
These factors are called linear factors because they have the form \(ax+by+c\).
Example 1 – factorising quadratics when \(a=1\)
Factorise \(x^{2}+5x+6\).
Multiplying the first term in each bracket gives the term \(x^{2}\), so the first terms must be \(x\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=+6\), so:
\[m\times n=6\]
\(m\) and \(n\) are factors of \(+6\) and could therefore be \(+1\) and \(+6\), or \(+2\) and \(+3\). Their negative versions could also be factors, such as \(-1\) and \(-6\), since \((-1)\times (-6)=+6\).
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=5\), so:
\[m+n=5\]
Looking at the possible values for \(m\) and \(n\), we can pick out the ones that add up to \(5\). This would be \(+2\) and \(+3\).
Therefore, the factorised form of \(x^{2}+5x+6\) is \((x+2)(x+3)\).
Factorise \(x^{2}+9x+14\).
Multiplying the first term in each bracket gives the term \(x^{2}\), so the first terms must be \(x\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=14\), so:
\[m\times n=14\]
\(m\) and \(n\) are factors of \(+14\) and could therefore be \(+1\) and \(+14\), or \(+2\) and \(+7\). Their negative versions could also be factors.
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=9\), so:
\[m+n=9\]
Looking at the possible values for \(m\) and \(n\), we can pick out the ones that add up to \(9\). This would be \(+2\) and \(+7\).
Therefore, the factorised form of \(x^{2}+9x+14\) is \((x+2)(x+7)\).
Factorise \(y^{2}-7y+12\).
Multiplying the first term in each bracket gives the term \(y^{2}\), so the first terms must be \(y\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=+12\), so:
\[m\times n=12\]
\(m\) and \(n\) are factors of \(12\) and could therefore be \(+1\) and \(+12\), or \(+3\) and \(+4\). Their negative versions could also be factors.
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=-7\), so:
\[m+n=(-7)\]
Looking at the possible values for \(m\) and \(n\), we can pick out the ones that add up to \(-7\). This would be \(-3\) and \(-4\).
Therefore, the factorised form of \(y^{2}-7y+12\) is \((y-3)(y-4)\).
Factorise \(p^{2}-5p-14\).
Multiplying the first term in each bracket gives the term \(p^{2}\), so the first terms must be \(p\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=-14\), so:
\[m\times n=(-14)\]
\(m\) and \(n\) are factors of \(-14\) and could therefore be \(+1\) and \(-14\), \(-1\) and \(+14\), \(+2\) and \(-7\), or \(-2\) and \(+7\).
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=-5\), so:
\[m+n=(-5)\]
Looking at the possible values for \(m\) and \(n\), we can pick out the ones that add up to \(-5\). This would be \(+2\) and \(-7\).
Therefore, the factorised form of \(p^{2}-5p-14\) is \((p+2)(p-7)\).
Factorise \(a^{2}+6a-7\).
Multiplying the first term in each bracket gives the term \(a^{2}\), so the first terms must be \(a\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=-7\), so:
\[m\times n=(-7)\]
\(m\) and \(n\) are factors of \(-7\) and could therefore be \(-1\) and \(+7\), or \(+1\) and \(-7\).
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=+6\), so:
\[m+n=6\]
Looking at the possible values for \(m\) and \(n\), we can pick out the ones that add up to \(+6\). This would be \(-1\) and \(+7\).
Therefore, the factorised form of \(a^{2}+6a-7\) is \((a-1)(a+7)\).
Factorise \(a^{2}+3a+6\).
Multiplying the first term in each bracket gives the term \(a^{2}\), so the first terms must be \(a\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=+6\), so:
\[m\times n=6\]
\(m\) and \(n\) are factors of \(6\) and could therefore be \(+1\) and \(+6\), \(-1\) and \(-6\), \(+2\) and \(+3\) or \(-2\) and \(-3\).
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=+3\), so:
\[m+n=3\]
Looking at the possible values for \(m\) and \(n\), we discover than none of them add up to \(+3\). This means that we cannot factorise \(a^{2}+3a+6\). In other words, there are no real linear factors.
It is important to understand when this occurs, which you will learn about later.
Your turn – factorising quadratics when \(a=1\)
Factorise the following expressions.
\(x^{2}+10x+21\)
\(z^{2}+11z+18\)
\(x^{2}+5x-14\)
\(m^{2}-m-72\)
\(x^{2}+6x+9\)
\(a^{2}-15a+44\)
\(x^{2}-2x-24\)
\(y^{2}-10y+16\)
\(z^{2}+4z-60\)
\(n^{2}+6n-16\)
\(a^{2}+5a+10\)
\(s^{2}+2s-48\)
\(y^{2}+7y+19\)
\(x^{2}+16x+39\)
\(x^{2}-14x+45\)
\((x+3)(x+7)\)
\((z+2)(z+9)\)
\((x+7)(x-2)\)
\((m-9)(m+8)\)
\((x+3)(x+3)\) or \((x+3)^{2}\)
\((a-4)(a-11)\)
\((x-6)(x+4)\)
\((y-8)(y-2)\)
\((z+10)(z-6)\)
\((n+8)(n-2)\)
Has no real linear factors
\((s+8)(s-6)\)
Has no real linear factors
\((z+13)(z+3)\)
\((x-9)(x-5)\)
Factorising when \(a\neq1\)
Expressions of the type \(ax^{2}+bx+c\), where \(a\neq1\), can be factorised using a technique similar to what we have used for \(x^{2}+bx+c\).
When we factorise these types of quadratics, the coefficient of \(x\), in at least one bracket, will not equal \(1\).
The examples shown use the trial-and-error method to factorise the quadratics. When there are many factors, the process can get quite tedious. In these cases, we can use other methods. You will learn about them later.
Example 1 – factorising quadratics where \(a\neq1\)
Factorise \(2x^{2}+7x+6\).
Multiplying the first term in each bracket gives the term \(2x^{2}\). The factors of \(2x^{2}\) are \(+1\) and \(+2x\), and its negative versions.
\(m=3\) and \(n=2\) satisfies the condition \(b=+7\). In other cases, you might have to continue and try the remaining possibilities.
Substituting \(m=3\) and \(n=2\) into the expression, the final factorised expression is:
\[(2x+3)(x+2)\]
Factorise \(6x^{2}+7x+2\).
Multiplying the first term in each bracket gives the term \(6x^{2}\). The factors of \(6x^{2}\) are \(+x\) and \(+6x\), \(+2x\) and \(+3x\), and their negative versions.
Fortunately, the first one we tried worked! In other cases, you might have to continue and try each possibility.
Substituting \(m=1\) and \(n=2\) into expression 2, the final factorised expression is:
\[(2x+1)(3x+2)\]
Factorise \(2x^{2}-10x+12\).
At first, this looks like a case where \(a=2\), but \(2\) is a common factor of all terms in this expression. It can be taken out to get:
\[2(x^{2}-5x+6)\]
It is good practice to check whether there are any common factors first! This can save you a lot of time. Here, we have an expression where \(a=1\), so we can factorise using the first method we learned on this page.
Multiplying the first term in each bracket gives the term \(x^{2}\), so the first terms must be \(x\).
Multiplying the last term in each bracket gives the constant term \(c\). \(c=+6\), so:
\[m\times n=6\]
\(m\) and \(n\) are factors of \(+6\) and could therefore be \(+1\) and \(+6\), \(+2\) and \(+3\), or their negative versions.
The coefficient \(b\) is the sum of the last term in each bracket. In this case, \(b=-5\), so:
\[m+n=-5\]
Looking at the possible values for \(m\) and \(n\), we can pick out the ones that add up to \(-5\). This would be \(-3\) and \(-2\).
Therefore, the factorised form of \(2x^{2}-10x+12\) is \(2(x-3)(x-2)\).
Factorise \(6x^{2}+13x-8\).
Multiplying the first term in each bracket gives the term \(6x^{2}\). The factors of \(6x^{2}\) are \(+x\) and \(+6\), \(+2\) and \(+3\), and their negative versions.
\(m=-1\) and \(n=+8\) satisfies the condition that \(b=+13\). In other cases, you might have to continue and try the remaining possibilities.
Substituting \(m=-1\) and \(n=+8\) into the expression, the final factorised expression is:
\[(2x-1)(3x+8)\]
Factorise \(4x^{2}+4x+1\).
Multiplying the first term in each bracket gives the term \(4x^{2}\). The factors of \(4x^{2}\) are \(+1\) and \(+4\), \(+2\) and \(+2\), and their negative versions.
\(m=+1\) and \(n=+1\) satifies the condition that \(b=+4\).
Substituting \(m=+1\) and \(n=+8\) into the expression, the final factorised expression is:
\[(2x+1)(2x+1)\textrm{ or }(2x+1)^{2}\]
Your turn – factorising quadratics when \(a\neq 1\)
Factorise the following.
\(5x^{2}+13x+6\)
\(2x^{2}+x-15\)
\(3m^{2}-m-2\)
\(3y^{2}-10y+8\)
\(2a^{2}+11a+12\)
\(6x^{2}-11x+5\)
\((5x+3)(x+2)\)
\((2x-5)(x+3)\)
\((3m+2)(m-1)\)
\((3y-4)(y-2)\)
\((2a+3)(a+4)\)
\((6x-5)(x-1)\)
The discriminant
As you saw in one of the earlier examples, sometimes a quadratic does not have real linear factors. But, you don't really want to be completing time-consuming calculations only to find out that you cannot factorise the quadratic.
One way to avoid this is to first check whether it has any linear factors. We do this using the discriminant.
The discriminant, denoted \(\Delta\), for the general quadratic \(ax^{2}+bx+c\) is:
\[\Delta=b^{2}-4ac\]
If:
\[\begin{align*} \Delta & = 0\textrm{, there is one repeated linear factor}\\
\Delta & > 0\textrm{, there are two distinct linear factors}\\
\Delta & < 0\textrm{, there are no real linear factors}
\end{align*}\]
Graphically, a quadratic is a parabola (a U-shaped curve). The discriminant tells us how many times the parabola touches or intersects the \(x\)-axis. When \(\Delta=0\), the graph just touches the \(x\)-axis at just one point. When \(\Delta>0\), the graph touches the \(x\)-axis at two points. When \(\Delta<0\), the graph does not intersect the \(x\)-axis.
Example – using the discriminant
Factorise \(x^{2}+6x+12\).
We can use the discriminant to determine whether the quadratic has real linear factors.
For this quadratic, \(a=+1\), \(b=+6\) and \(c=+12\).
