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Stage 4 - Stage 5

Monic quadratic trinomials


We have already learnt how to expand the product of two binomials using the FOIL technique, the result of which is called a quadratic, which is any algebraic expression where only one type of variable exists but the highest power of that variable is $2$2. For example, $2x^2-5x+1$2x25x+1 is a quadratic, and so is $-x^2-47$x247. Now we're going to learn how to reverse that expansion process and factorise to get back to our original binomials. We're going to focus on a special type of quadratic called a monic quadratic, which just means that the coefficient for the $x^2$x2 part is always going to be $1$1


Now we know that $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$(x+a)(x+b)=x2+(a+b)x+ab

So to factorise a monic quadratic expression like $x^2+Px+Q$x2+Px+Q,

1. Find two numbers, $a$a and $b$b, that have a product of $Q$Q and a sum of $P$P

  • It's usually a lot easier to look at $Q$Q and start listing out its factors first rather than first finding the numbers that add to make $P$P!

2. The factorised form should be $\left(x+a\right)\left(x+b\right)$(x+a)(x+b)


Watch Out!
  • If $Q$Q is positive and $P$P is positive, then $a$a and $b$b are positive
  • If $Q$Q is positive and $P$P is negative, then $a$a and $b$b are negative
  • If $Q$Q is negative, then one of the two numbers is positive and the other is negative, regardless of $P$P


Cross method

Another way to factorise quadratics is called the cross method. This method can be used even for non-monic quadratics, so let's have a look at how it works! Let's take the example $x^2+7x+6$x2+7x+6. We know the answer must be of the form $\left(x+a\right)\left(x+b\right)$(x+a)(x+b), as only this format will give us the first term $x^2$x2, since the only factor pair of $x^2$x2 is $x$x & $x$x (Technically $x^2$x2 & $1$1 is another pair, but it would not help us factorise!). Then our job is just to find what $a$a and $b$b are. We know they must multiply together to give us $6$6, so let's take a look at the factor pairs of $6$6, which are $2$2 & $3$3 or $1$1 & $6$6.

The cross method works by trial and error, and we start by drawing a cross, with a possible factor pair of the first $x^2$x2 term on the left, and a possible factor pair of $6$6 on the right. Let's try the pair $2$2 & $3$3 for now:

Then we follow along each arrow to make a product of the numbers linked by them. For example the blue arrow gives us $3x$3x while the green one gives us $2x$2x. Together they have a sum of $5x$5x, but our goal is to reach $7x$7x, the middle term of the quadratic. That means $2$2 & $3$3 must not work. Let us try $6$6 & $1$1 then:

Hooray! This configuration does indeed give us the sum we're looking for, which is $7x$7x. To read off the correct factorised form of this quadratic then, simply put each circled pair in brackets and multiply them together. For example here we can factorise $x^2+7x+6$x2+7x+6 to become $\left(x+1\right)\left(x+6\right)$(x+1)(x+6).



Question 1

Factorise $x^2+15x+56$x2+15x+56

Think about whether $a$a and $b$b are positive or negative here


$a+b=15$a+b=15 and $a\times b=56$a×b=56

$a$a and $b$b must both then be positive

If we look at the factors of $56$56, we get these pairs:

$1$1, $56$56

$2$2, $28$28

$4$4, $14$14

$7$7, $8$8

The only pair with a sum of $15$15 is the last pair, so $a$a and $b$b must equal $7$7 and $8$8.

So $x^2+15x+56=\left(x+7\right)\left(x+8\right)$x2+15x+56=(x+7)(x+8)


Question 2

Use the cross method to solve $x^2+2x-3$x2+2x3.

Think about how there's two different factor pairs for $-3$3 as it is a negative number


$-3$3 has the factor pairs $1$1 & $-3$3 and $-1$1 & $3$3. Let us try the first pair first:

This does not give us the answer we want which is $2x$2x. Let's then try the other pair:

This gives us the correct sum so the factorised form must be $\left(x-1\right)\left(x+3\right)$(x1)(x+3).

Question 3

Factorise $x^2-2x-8$x22x8.


Question 2

Factorise $44-15x+x^2$4415x+x2.


Question 5

Factorise the expression completely by first taking out a common factor:



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