## Quadratic expressions

A quadratic expression is an expression in which the highest power of unknown is 2. e.g.

- x
^{2}+ 9x + 14 - x
^{2}+ 14 - x
^{2}– 9

## Factorising quadratic expressions by grouping

The method of factorisation by grouping can be extended even to quadratic expressions. In the previous exercise there were some quadratic expressions we factorised which were however in different form from what we are going to look at right now.

**Example**

Factorise the following completely

- x
^{2}+ 9x + 14

*This expression has 3 terms and we need 4 terms to factorise it by grouping pairs. So we split the middle term into 2 terms as follows.*

*first we multiply the end terms, x ^{2} and 14 in this case to get 14x^{2}*

*Then we list the factors of 14x ^{2} and find among them, 2 factors whose sum is 9x, the middle term.*

14x^{2} =

- x × 14x [add these and you get 15x.]
- 2x × 7x [add these and you get 9x.]

*Therefore the factors we need are 2x and 7x. Go back to the oroginal expression and replace 9x with these 2 factors.*

= x^{2} + 9x + 14

= x^{2} + 2x + 7x + 14

*then factorise by grouping*

= x(x + 2) + 7(x + 2)

= (x + 2)(x + 7)

**More Examples**

Factorise the following completely

- a
^{2}+ 8a + 15

*Multiply 15 and a*^{2}to get 15a^{2}

*Find the factors of 15a*^{2}whose sum is 8a

*Let me tabulate the process to make it easier to understand.*

Factors | Sum | Valid |
---|---|---|

a ×15a | 16a | No |

3a × 5a | 8a | Yes |

*Replace 8a in the original expression by 3a and 5a*

= a^{2} + 8a + 15

= a^{2} + 5a + 3a + 15

= a(a + 5) + 3(a + 5)

= (a + 5)(a + 3)

- b
^{2}– 7b + 10

*Lets find factors of 10b*^{2}whose sum is -7b

Factors | Sum | Valid |
---|---|---|

2b ×5b | 7b | No |

-2b ×-5b | -7b | Yes |

= b^{2} – 7b + 10

= b^{2} – 2b – 5b + 10

= b(b – 2) – 5(b – 2)

= (b – 2)(b – 5)

**However with practice you will soon master the art of finding the factors without tabulating anything.**

- c
^{2}+ 4c – 21

*Multiply c*^{2}by -21 to get -21c^{2}and then find the factors of -21c^{2}, whose sum is 4c

*Thats 7c and -3c.*

= c^{2}+ 7c – 3c – 21

= c(c + 7) – 3(c + 7)

= (c + 7)(c – 3)

- d
^{2}– 5d – 14

*Find the factors of -14d*^{2}whose sum is -5d

*Thats -7d and 2d.*

= d^{2}– 7d + 2d – 14

= d(d – 7) + 2(d – 7)

= (d – 7)(d + 2)

**As you can see, the process is the same, its only the numbers and factors that keep changing, so the next couple of examples do not have explanations.**

- e
^{2}+ 2e – 8

*e*^{2}by -8 = -8e^{2}. So find the factors of -8e^{2}whose sum is 2e

*Thats -2e and 4e*

= e^{2}-2e + 4e – 8

= e(e – 2) + 4(e – 2)

= (e + 4)(e – 2)

- w
^{2}+ 5w + 6

*w*^{2}by 6 = 6w^{2}. Find the factors of 6w^{2}whose sum is 5w

*Thats 2w and 3w*

= w^{2}+ 2w + 3w + 6

= w(w + 2) + 3(w + 2)

= (w + 2)(w + 3)

- x
^{2}+ 5x – 6

*x*^{2}by -6 = -6x^{2}. Find the factors of -6x^{2}whose sum is 5x

*Thats -x and 6x*

= x^{2}-x + 6x – 6

= x(x – 1) + 6(x – 1)

= (x – 1)(x + 6)

- y
^{2}– 5y + 6

*y*^{2}by 6 = 6y^{2}. Find the factors of 6y^{2}whose sum is -5y

*Thats -2y and -3y*

= y^{2}– 2y – 3y + 6

= y(y – 2) -3(y – 2)

= (y – 3)(y – 2)

## Difference of 2 squares

A perfect square is a number that has an exact square root, e.g: 1, 4, 9, 16 are perfect squares and their square roots are 1, 2, 3, 4 respectively. We can also have perfect squares among algebraic terms, e.g x^{2} is a perfect square whose square root is x.

Now when we talk about difference of 2 squares we mean a quadratic equation like this (x^{2} – 4) in which we have 2 squares subtracting each other.

Take note that:

- (x
^{2}+ 4) is not difference of 2 squares because those squares are not subtracting each other. - (x
^{2}– 4x) is not difference of 2 squares because 4x is not a perfect square due to the presence of x. - (x
^{2}– 4y^{2}) is difference of 2 squares.

**Example**

Factorise the following:

- x
^{2}– 4

*Find the square roots of both squares and express them in 2 brackets, first as a sum and then as a difference*

= (x + 2)(x – 2)

- r
^{2}– 25

= (r + 5)(r – 5)

- x
^{2}– y^{2}

= (x + y)(x – y)

- x
^{2}– 36y^{2}

= (x – 6y)(x + 6y)

- a
^{2}b^{2}– 9c^{2}

= (ab + 3c)(ab – 3c)

#### Disguised difference of 2 squares

Sometimes the squares can be disguised until you first pull out a common factor.

**Examples**

- 2b
^{2}– 18

*As it is its not difference of 2 squares, until we pull out the common factor 2.*

= 2(b^{2}– 9)

= 2(b + 3)(b – 3)

- ax
^{2}– ab^{2}

= a(x^{2}– b^{2})

= a(x + b)(x – b)

## Other quadratic expressions with 2 terms

**Examples**

- x
^{2}– 4x

= x(x – 4)

- m
^{2}+ 10m

= m(m + 10)