Factorising Quadratics : Theory

• • When we factorise a quadratic equation, we split the quadratic into 2 brackets (double bracket) 
  • If we were to multiply the 2 brackets, they would give us the quadratic equation
• • Quadratics are written in the form ax² + bx + c 
  • In order to factorise a quadratic, we need to find 2 numbers / terms that add to make ‘b’ and multiply to make ‘c’ 
  • We can do this, by finding factors of ‘c’
• • Quadratic can also be in the form ax² - bx + c 
  • To solve this, we need to realise that multiplying 2 negative numbers results in a positive number 
  • So, 2 negative factors of ‘c’ , will add to make ‘b’
• • Quadratics can also be in the form ax² - bx - c  or ax² + bx - c
  • In this case, one of the factors of c will be positive, and the other will be negative
• • Quadratics can also be in the form x² - c
  • In this case, c will always be a square number 
  • We call this the difference between 2 squares 
  • In quadratics in this form, there are middle terms (bx), because the 2 factors of c are the same, but one is positive and the other is negative 
  • These 2 factors will cancel each other out
• • Sometimes, quadratics will not be written in the form ax² + bx + c, which means we cannot split the quadratic into 2 brackets 
  • Instead, if they are written in forms such as ax² - bx , we factorise them using a single bracket 
  • The highest common factor between the 2 terms goes on the outside of the single bracket 
  • e.g. x(ax - b)
• • Sometimes, a quadratic will be written in the form ax² + bx + c , where the coefficient of x² , which is a, is bigger than 1 
  • To solve this, we would find the factors of c, and then multiply only one of the factors for all of the possible combinations of factors of c 
  • We would then select the factors that add together to make b

  
• • Sometimes, you can get a quadratic in the form ax² + bx + c, where the ax² can split into 2 ax terms 
  • However, the 2 ax terms could both be greater than 1 
  • To solve this, we would find the 2 factors of c 
  • We would multiply one of the factors of all the possible combinations of factors for c, by one of the coefficients of the ax² 
  • We would do the same fore the other factors, and multiply them by the other coefficient of the ax² 
  • We would chose the 2 factors that add to give b

  
• • Sometimes, we can get a quadratic in the form ax² - c
  • This is in the format of the difference of two squares, however the coefficient of x may be greater than 1 
  • To solve this, we split the quadratic into 2 brackets - each bracket will start with the square root of the coefficient of x² - i.e the number that multiplies by itself to form the coefficient of x² 
  • And in one bracket, the positive square root of c will be at the end, and in the other, the negative square root of c will be at the end 

  
• • Sometimes, we can get a quadratic in the form ax² - by² , where a and b are greater than 1 
  • This is the difference of 2 squares, and a is the coefficient of x², and b is the coefficient of y²
  • To solve this, we factorise the quadratic 2 brackets, and we place the square root of ax² at the start of each 
  • Then in one of the brackets, we add the square root of ax² to the square root of by²  
  • Then, in the other bracket, we do the square root of ax² minus the square root of by²

  
• • If you get a quadratic in the form ax² - c , and the coefficient of x² is not a square number, you have to find the HCF between a and c 
  • Initially we use a single bracket 
  • However, once you find the term in the bracket e.g. b(x² - c), you can factorise this term, if c is also a square number, by splitting the term into 2 brackets - this term is simply the difference of 2 squares
  • So the positive square root of c would be in one bracket, and the negative square root of c would be in the other

  
• • If we have the quadratic In the form x⁴ - c, we can factorise it into 2 brackets if c is also a square number - remember that x² x x² = x⁴ 
  • So in 2 brackets, we would have (x² + √c) (x² - √c) 
  • If √c results in another square number, we can factorise the second bracket, as again, it would be the difference of 2 squares 
  • We would simply leave the first bracket the same, and split the second bracket into 2 brackets

  E.g. (x² + √c) (x + √d) (x² + √d)

  
• This is what to do in order to factorise 2 algebraic terms to the power of different values, where the term on the left has a greater power than the term on the right

  
• This is what to do in order to factorise 2 algebraic terms to the power of different values, where the term on the right has a greater power than the term on the left

  
• • To factorise ax²y⁶ - c , you would split the expression into 2 brackets
  • Remember that x multiplied by x =x²
  • And y³ x y³ = y⁶

  
• • This is what to do when simplifying 2 brackets that are both squared - you don't need to expand the bracket, as the square root of a bracket to the power of 2 is just the term in the bracket
  • Remember to set up the terms in the form of the difference of two squares

  
• • This is what to do when factorising a square number subtract a squared bracket 
  • Remember to set up the terms in the form of the difference of 2 squares

  
• • This is what to do when factorising 2 fractions that are made up of square numbers 
  • Remember to set up the terms in the form of the difference between 2 squares