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To understand how to factorise quadratic expressions, it is important you know how to expand double brackets, check the expanding part 2 page to learn this first, as knowing this beforehand will allow you to understand factorisation better.
Remember that when expanding two brackets, lets say (y + 3)(y - 5) we would multiply the first term in the first bracket with both terms in the second bracket (one by one) and then multiply the second term in the first bracket with both terms in the second bracket (one by one) then add the products.
y x y = y2 y x -5 = -5y
3 x y = 3y3 x -5 = -15
Adding the products gives us y2 - 5y + 3y - 15 = y2 - 2y - 15
Thus, (y + 3)(y - 5) = y2 - 2y - 15
What if the question was factorise y2 - 2y - 15, in other words how do we get from y2 -2y -15 to (y + 3)(y -5)? Well, we factorise the expression.
Question 1: Factorise y2 - 2y - 15
So lets assume we are given the expression y2 - 2y + 15 and this is the expression we need to factorise.Whenever we have a quadratic expression, remember a quadratic expression is an expression in the form of ax2 + bx + c. The only difference here is that the question has used the letter y instead of the letter x.
In our case a = 1, b = -2, c = -15
When we factorise a quadratic expression that can be factorised, there will always be two brackets being multiplied by each other.
So y2 - 2y - 15 = ( )( ).
The question is what do we put in those brackets, so that when we expand the double brackets we get back our quadratic expression.
Lets think about it logically, to get the term y2, we need to have y x y when expanding the brackets. So there HAS TO be a y term in each bracket to get the y2 term.
So currently we have
(y )(y )
How do we get the -2y and -15 term?
Well, first of all, lets remember, what does a, b and c equal.
In the quadratic equation y2 - 2y - 15, so a = 1, b = -2, c = -15.
a x c = -15
Given that ac = -15 (remember we can write ac = a x c)
We need to find two numbers that multiply to get -15 AND can be added or subtracted to get -2, which equals b. YOU NEED TO UNDERSTAND THIS, READ IT AGAIN IF YOU DO NOT.
These two numbers are 3 and -5.
Lets check, if we choose, does -3 x 5 = 15?
YES it does
Can we add or subtract 3 and -5 to get -2? Yes we can add 3 and -5 together to get -2. 3 + -5 = -2
So now we have the two numbers that multiply to get -15 AND can be added or subtracted to get -2.
Lets insert these two numbers into
(y )(y ) to get
(y + 3)(y - 5) We now have our expression factorised, and thus this is the answer.
Lets think about why we chose 3 and -5 because the same concept can be applied to other questions. We understand that to get the y2 term, that there needs to be a y term in either bracket.
Lets understand this better by expanding the brackets
y x y = y2 y x -5 = -5y 3 x y = 3y3 x -5 = -15
Can you see why it is important that we choose two numbers which add or subtract to get -2, because the coefficient of y in our quadratic equation is -2. The two numbers that we choose will each be multiplied with y which when added will need to give -2y, as shown by the products above. However, we also need to take into consideration that the two numbers multiply to get -15, as the constant in our quadratic equation -15.
Question 2: Factorise y2 - 6x + 8
So we know that this is a quadratic, as y2 - 6y - 8 is in the form of ax2 + bx + c, where a = 1, b = -6 and c = -8, the only difference is that this question has used the letter y instead of the letter x.
As this is a quadratic WE KNOW THAT there would be two brackets, AND, as there is a y2, we already know that there has to be a y term in both brackets, that is how we get y x y = y2.
So we currently have
(y )(y )
Now lets find two numbers that can be added or subtracted get -6 and multiply to get ac. Remember a x c = ac = 1 x 8 = 8.
The two numbers are -4 and -2, if we add -4 and -2 we get -6, and when we multiply -4 and -2 we get 8, as shown below.
-4 + -2 = -4 - 2 = -6-4 x -2 = 8 (Remember that a negative number multiplied by a negative number is a positive number)
so lets insert the two numbers into
(y )(y )
to get
(y - 4)(y - 2)
Lets check whether we get y2 - 6 + 8 when we expand (y - 4)(y - 2).
y x y = y2 y x -2 = -2yy x -4 = -4y-4 x -2 = 8
Add the products to get
y2 -2y -4y + 8 = y2 -6y + 8
Remember that when expanding two brackets, lets say (y + 3)(y - 5) we would multiply the first term in the first bracket with both terms in the second bracket (one by one) and then multiply the second term in the first bracket with both terms in the second bracket (one by one) then add the products.
y x y = y2 y x -5 = -5y
3 x y = 3y3 x -5 = -15
Adding the products gives us y2 - 5y + 3y - 15 = y2 - 2y - 15
Thus, (y + 3)(y - 5) = y2 - 2y - 15
What if the question was factorise y2 - 2y - 15, in other words how do we get from y2 -2y -15 to (y + 3)(y -5)? Well, we factorise the expression.
Question 1: Factorise y2 - 2y - 15
So lets assume we are given the expression y2 - 2y + 15 and this is the expression we need to factorise.Whenever we have a quadratic expression, remember a quadratic expression is an expression in the form of ax2 + bx + c. The only difference here is that the question has used the letter y instead of the letter x.
In our case a = 1, b = -2, c = -15
When we factorise a quadratic expression that can be factorised, there will always be two brackets being multiplied by each other.
So y2 - 2y - 15 = ( )( ).
The question is what do we put in those brackets, so that when we expand the double brackets we get back our quadratic expression.
Lets think about it logically, to get the term y2, we need to have y x y when expanding the brackets. So there HAS TO be a y term in each bracket to get the y2 term.
So currently we have
(y )(y )
How do we get the -2y and -15 term?
Well, first of all, lets remember, what does a, b and c equal.
In the quadratic equation y2 - 2y - 15, so a = 1, b = -2, c = -15.
a x c = -15
Given that ac = -15 (remember we can write ac = a x c)
We need to find two numbers that multiply to get -15 AND can be added or subtracted to get -2, which equals b. YOU NEED TO UNDERSTAND THIS, READ IT AGAIN IF YOU DO NOT.
These two numbers are 3 and -5.
Lets check, if we choose, does -3 x 5 = 15?
YES it does
Can we add or subtract 3 and -5 to get -2? Yes we can add 3 and -5 together to get -2. 3 + -5 = -2
So now we have the two numbers that multiply to get -15 AND can be added or subtracted to get -2.
Lets insert these two numbers into
(y )(y ) to get
(y + 3)(y - 5) We now have our expression factorised, and thus this is the answer.
Lets think about why we chose 3 and -5 because the same concept can be applied to other questions. We understand that to get the y2 term, that there needs to be a y term in either bracket.
Lets understand this better by expanding the brackets
y x y = y2 y x -5 = -5y 3 x y = 3y3 x -5 = -15
Can you see why it is important that we choose two numbers which add or subtract to get -2, because the coefficient of y in our quadratic equation is -2. The two numbers that we choose will each be multiplied with y which when added will need to give -2y, as shown by the products above. However, we also need to take into consideration that the two numbers multiply to get -15, as the constant in our quadratic equation -15.
Question 2: Factorise y2 - 6x + 8
So we know that this is a quadratic, as y2 - 6y - 8 is in the form of ax2 + bx + c, where a = 1, b = -6 and c = -8, the only difference is that this question has used the letter y instead of the letter x.
As this is a quadratic WE KNOW THAT there would be two brackets, AND, as there is a y2, we already know that there has to be a y term in both brackets, that is how we get y x y = y2.
So we currently have
(y )(y )
Now lets find two numbers that can be added or subtracted get -6 and multiply to get ac. Remember a x c = ac = 1 x 8 = 8.
The two numbers are -4 and -2, if we add -4 and -2 we get -6, and when we multiply -4 and -2 we get 8, as shown below.
-4 + -2 = -4 - 2 = -6-4 x -2 = 8 (Remember that a negative number multiplied by a negative number is a positive number)
so lets insert the two numbers into
(y )(y )
to get
(y - 4)(y - 2)
Lets check whether we get y2 - 6 + 8 when we expand (y - 4)(y - 2).
y x y = y2 y x -2 = -2yy x -4 = -4y-4 x -2 = 8
Add the products to get
y2 -2y -4y + 8 = y2 -6y + 8
Why is factorisation so important?
Why is factorisation so important?
The importance of factorisation is that when we see the expression factorised we can find the values of x which make the value of the expression equal to 0, you will understand what this means better below. IT IS NOT HARD TO UNDERSTAND.
e.g. Lets say we wanted to find the value of y which makes y2 - 6 + 8 = 0
How do we do this? Well the whole point of factorisation is to write the expression in a different way, so we can learn more about the expression.
y2 - 6 + 8 = (y - 4)(y - 2),
so the question asks us to solve y2 - 6 + 8 = 0
We can write y2 - 6 + 8 = (y -4)(y - 2) = 0
So by looking at the factorised expression, we can solve and find the values of y which makes the expression equal 0.
(y -4)(y -2) = (y - 4) x (y - 2) = 0
If one bracket equals 0, then automatically 0 multiplied the value of the other bracket will give you 0.
So, here we have 2 brackets, either one can equal 0, and they WILL ONLY EQUAL 0 when y = 4, or y = 2.
Lets see this,
When y = 4 ONLY, the first bracket equals 0
(4 - 4)(4 - 2) = (0)(2) = 0 x 2 = 0
When y = 2 ONLY, the second bracket equals 0
(2 - 4)(2 - 2) = (-2)(0) = -2 x 0 = 0
So there are two values of y which solve y2 - 6 + 8 = 0 and these solutions are y = 4 AND y = 2
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