quadratic functions...factorization is the opposite of expanding brackets. after factorization of a...

QUADRATIC FUNCTIONS Factorizing Quadratic Expressions

1 | P a g e

Factorization is the opposite of expanding brackets. After factorization of a

quadratic expression has taken place, we will end up with a binomial expression.

Therefore, 𝑎𝑥2 + 𝑏𝑥 + 𝑐 becomes (𝑥 + 𝑑)(𝑥 + 𝑒)

PRIOR KNOWLEDGE: OPERATING WITH INTEGERS

Can you think of two numbers that:

1. give a sum of 5 and a product of 6? _______ and _______

2. give a sum of −5 and product of 6? _______ and _______

3. give a sum of 1 and a product of −6? _______ and _______

4. give a sum of −1 and a product of −6? _______ and _______


6. give a sum of −7 and a product of 12? _______ and _______

7. give a sum of 1 and a product of −12? _______ and _______




Mastering this skill, will help you to quickly factorize quadratic expressions,

especially when the coefficient of the 𝑥2 term is 1.


2 | P a g e

FACTORIZING QUADRATICS (when the coefficient of the 𝒙𝟐 is 1)

Factorize the following

Remember, 𝑎𝑥2 + 𝑏𝑥 + 𝑐 will become (𝑥 + 𝑑)(𝑥 + 𝑒)

1. 𝑥2 + 5𝑥 + 6

(𝑥 ) (𝑥 ) we need two numbers that give a sum of +5 and

a product of +6

(𝑥 + 2) (𝑥 + 3)

2. 𝑥2 − 5𝑥 + 6

(𝑥 ) (𝑥 ) we need two numbers that give a sum of −5

and a product of +6

(𝑥 − 2) (𝑥 − 3)

3. 𝑥2 + 𝑥 − 6

(𝑥 ) (𝑥 ) we need two numbers that give a sum of +1 and

a product of −6

(𝑥 − 2) (𝑥 + 3)

4. 𝑥2 − 𝑥 − 6


and a product of −6

(𝑥 + 2) (𝑥 − 3)

5. 𝑥2 + 7𝑥 + 12

(𝑥 ) (𝑥 ) we need two numbers that give a sum of +7

and a product of +12

(𝑥 + 3) (𝑥 + 4)


3 | P a g e

6. 𝑥2 − 7𝑥 + 12



(𝑥 − 3) (𝑥 − 4)

7. 𝑥2 + 𝑥 − 12



(𝑥 − 3) (𝑥 + 4)

8. 𝑥2 − 𝑥 − 12



(𝑥 + 3) (𝑥 − 4)

9. 𝑥2 + 2𝑥 + 15



(𝑥 + 5) (𝑥 − 3)

10. 𝑥2 − 2𝑥 − 15



(𝑥 + 3) (𝑥 − 5)

11. 𝑥2 − 5𝑥 − 24 LET’S SEE IF YOU GET IT



(𝑥 + ) (𝑥 − )


4 | P a g e

FACTORIZING QUADRATICS (when the coefficient of the 𝒙𝟐 is greater than 1)

Factorize the following

Remember, 𝑎𝑥2 + 𝑏𝑥 + 𝑐 will become (𝑥 + 𝑑)(𝑥 + 𝑒)

1. 𝟐𝒙𝟐 + 𝟕𝒙 + 𝟔

𝑎 = 2, 𝑏 = 7 and 𝑐 = 6

𝑎𝑐 = 2 x 6

𝑎𝑐 = 12

We need two numbers that will give a sum of +7 and a product of +12.

These are +3 and +4. These factors will be used to replace the middle term, 7𝑥.

𝟐𝒙𝟐 + 𝟕𝒙 + 𝟔

𝟐𝒙𝟐 + 𝟒𝒙 + 𝟑𝒙 + 𝟔 We now group in terms of two.

(𝟐𝒙𝟐 + 𝟒𝒙) + (𝟑𝒙 + 𝟔) Factorize each grouped expression

𝟐𝒙(𝒙 + 𝟐) + 𝟑(𝒙 + 𝟐) Place one of each factor in brackets

(𝟐𝒙 + 𝟑)(𝒙 + 𝟐) This is your answer

If we were asked to solve the quadratic, we would now equate each factor to

0 then solve for 𝒙.

So 2𝑥 + 3 = 0 and 𝑥 + 2 = 0

When 2𝑥 + 3 = 0 When 𝑥 + 2 = 0

Then 2𝑥 = 0 − 3 Then 𝑥 = 0 − 2

2𝑥 = −3 ∴ 𝑥 = −2

2𝑥

2=

−3

2

∴ 𝑥 = −3

2

The ROOTS of the quadratic are 𝑥 = −3

2 and 𝑥 = −2.


5 | P a g e

The graph below shows the parabola curve for the quadratic expression,

𝟐𝒙𝟐 + 𝟕𝒙 + 𝟔 . See that it crosses the 𝑥 − 𝑎𝑥𝑖𝑠 two places. One at −3

2 and the

other at −2.

Therefore, the root of a quadratic is found where the curve crosses the 𝑥 − 𝑎𝑥𝑖𝑠.


6 | P a g e

2. 𝟐𝒙𝟐 − 𝟕𝒙 + 𝟔

𝑎 = 2, 𝑏 = −7 and 𝑐 = 6

𝑎𝑐 = 2 x 6

𝑎𝑐 = 12

We need two numbers that will give a sum of −7 and a product of +12.

These are −3 and −4. These factors will be used to replace the middle term,

−7𝑥. These must be so placed that they are factors of the first and fourth

term. That is, the first should be able to go into the second and the third into

the fourth.

𝑆𝑜, 2𝑥2 − 𝟕𝒙 + 6 becomes

2𝑥2 − 𝟒𝒙 − 𝟑𝒙 + 6 We now group in terms of two

(2𝑥2 − 4𝑥) + (−3𝑥 + 6) Factorize each grouped expression

2𝑥(𝑥 − 2) + −3(𝑥 − 2) Simplify signs in the middle

2𝑥(𝑥 − 2) − 3(𝑥 − 2) Place one of each factor in brackets

(2𝑥 − 3)(𝑥 − 4) This is your answer

If we were asked to solve the quadratic, we would now equate each factor to

0 then solve for 𝒙.

So 2𝑥 − 3 = 0 and 𝑥 − 2 = 0

When 2𝑥 − 3 = 0 When 𝑥 − 2 = 0

Then 2𝑥 = 0 + 3 Then 𝑥 = 0 + 2

2𝑥 = 3 ∴ 𝑥 = 2

2𝑥

2=

3

2

∴ 𝑥 = 3

2 The ROOTS are 𝑥 =

3

2 and 𝑥 = 2


7 | P a g e


𝟐𝒙𝟐 − 𝟕𝒙 + 𝟔 . See that it crosses the 𝑥 − 𝑎𝑥𝑖𝑠 two places. One at 3

2 and the

other at +2.


8 | P a g e

3. −𝟑𝒙𝟐 − 𝟕𝒙 + 𝟔

𝑎 = −3, 𝑏 = −7 and 𝑐 = 6

𝑎𝑐 = −3 x 6

𝑎𝑐 = −18

We need two numbers that multiply to give a sum of −7 and a product of −18

These are 2 and − 9. These factors will be used to replace the middle term,

−7𝑥.

−3𝑥2 − 7𝑥 + 6

−3𝑥2 − 9𝑥 + 2𝑥 + 6 We now group in terms of two

(−3𝑥2 − 9𝑥) + (2𝑥 + 6) Factorize each grouped expression

−3𝑥(𝑥 + 3) + 2(𝑥 + 3) Place one of each factor in brackets

(−3𝑥 + 2)(𝑥 + 3)

Calculating the roots of the above quadratic:

−3𝑥 + 2 = 0 and 𝑥 + 3 = 0

When −3𝑥 + 2 = 0 When 𝑥 + 3 = 0

Then −3𝑥 = 0 − 2 Then 𝑥 = 0 − 3

−3𝑥 = −2 ∴ 𝑥 = −3

−3𝑥

−3=

−2

−3

∴ 𝑥 = 2

3

The ROOTS of the quadratic are 𝑥 = 2

3 and 𝑥 = −3


9 | P a g e


−𝟑𝒙𝟐 − 𝟕𝒙 + 𝟔 . See that it crosses the 𝑥 − 𝑎𝑥𝑖𝑠 two places. One at 3

2 and the

other at +2.


10 | P a g e

THE PERFECT SQUARE (coefficient of the 𝒙𝟐 term is 1)

An expression such as, 𝑥2 + 2𝑥 + 1 is considered a perfect square because the

value of the 𝑐 term is found by taking half of the 𝑏 term then squaring it.

So, 𝑥2 + 2𝑥 + (2

2)

2 = 𝑥2 + 2𝑥 + 12

= 𝑥2 + 2𝑥 + 1

Now let’s consider multiplying the following binomial expression, (𝑥 + 1)2.

Recall that (𝑥 + 1)2 means (𝑥 + 1)(𝑥 + 1)

So expanding would be done by doing 𝑥(𝑥 + 1) + 1(𝑥 + 1)

𝑥2 + 𝑥 + 𝑥 + 1

𝑥2 + 2𝑥 + 1

Recall that factorization is the opposite of expanding bracket. Therefore, when we

factorize 𝑥2 + 2𝑥 + 1 we should get (𝑥 + 1)(𝑥 + 1). But they have the same

factor, so we write

𝒙𝟐 + 𝟐𝒙 + 𝟏 = (𝒙 + 𝟏)𝟐.

If 𝑥2 + 4𝑥 + 𝑐 is a perfect square, what is the value of 𝑐 ?

Recall that 𝑐 is found by taking half of the 𝑏 term then squaring it. So, 𝑐 = 4

Therefore, the quadratic expression would be 𝑥2 + 4𝑥 + 4.

Since 𝑥2 + 4𝑥 + 8 is a perfect square when factorize it will become, (𝑥 + 2)2

To factorize the expression, it will be (𝑥 + ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑏 𝑡𝑒𝑟𝑚) all squared.

If 𝑥2 + 𝑏𝑥 + 25 is a perfect square what is the value of 𝑏?

We would have to take the square root of 25, which is ∓5 then multiply it by

2.

So the value of 𝑏 would be 10.

The perfect square would be 𝑥2 + 10𝑥 + 25 or 𝑥2 − 10𝑥 + 25


11 | P a g e

Factorize the following perfect square expressions

𝑎𝑥2 + 𝑏𝑥 + 𝑐 will become (𝑥 + 𝑑)2, where 𝑑 is half of the 𝑏 term

1. 𝑥2 + 6𝑥 + 9

(𝑥 + 3)2

2. 𝑥2 − 6𝑥 + 9

(𝑥 − 3)2

3. 𝑥2 + 12𝑥 + 36

(𝑥 + 6)2

4. 𝑥2 − 12𝑥 + 36

(𝑥 − 6)2

5. 𝑥2 − 5𝑥 +25

4 Recall that half of −5 can be written as

−5

2

(𝑥 −5

2)

2 When squared, (

−5

2)

2means

−5

2 x

−5

2 =

25

4

6. 𝑥2 + 3𝑥 +9

4

(𝑥 +3

2)

2


12 | P a g e

THE DIFFERENCE OF TWO SQUARES

The difference of two square terms would look like this, 𝑎2 − 𝑏2.

Let’s consider finding the value of 42 − 92.

42 − 92 = 16 − 81

= −65

Let’s now consider finding the value of (4 + 9) (4 −9)

(4 + 9) (4 −9) (4 + 9) (4 −9) binomial expansion

(13) (−5) 4(4 − 9) + 9(4 − 9)

13 x −5 16 −36 + 36 − 81

−65 16 + 0 − 81

16 − 81 which is the same as 42 − 92

−65

Therefore, 𝑎2 − 𝑏2, when factorized becomes (𝑎 + 𝑏)(𝑎 − 𝑏)


13 | P a g e

Factorize the following expressions (Difference of two squares)

1. 𝑥2 − 𝑦2 (𝑥 + 𝑦)( 𝑥 − 𝑦)

2. 𝑒2 − 1 can also be written as 𝒆𝟐 − 𝟏𝟐

(𝑒 + 1)(𝑒 − 1)

3. 𝑚2 − 4 can also be written as 𝒎𝟐 − 𝟐𝟐

(𝑚 + 2)(𝑚 − 2)

4. 4𝑎2 − 9 can also be written as 𝟐𝟐𝒂𝟐 − 𝟑𝟐

(2𝑎 + 3)(2𝑎 − 3)

5. 25 − 16𝑝2 can also be written as 𝟓𝟐 − 𝟒𝟐𝒑𝟐

(5 + 4𝑝)(5 − 4𝑝)

quadratic functions...factorization is the opposite of expanding brackets. after factorization of a...

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