## Key Questions

• Good question .

basically what are asking me to factor is :

$a {x}^{2} + b x + c$

here is an eg.

4x^2 – 15x + 9.

First you need to make column like this

SUM = COEFFICIENT OF x =( -15)
product = (COEFFICIENT of x^2 $\cdot$ COEFFICIENT of constant)= 36
Now you have to find factors which have a sum of -15 and a product of 36

THE FACTORS ARE (-12), (-3)

So now what we do is split the middle term in the equation and factor it out

4x^2 – 12x– 3x +9.
4x(x – 3)–3 (x-3).

(4x–3 )(x – 3).

HENCE FACTORED

f(4x^2 – 15x + 9.) = (4x–3 )(x – 3)

• Example 1

${x}^{2} - x - 6 = \left(x + 2\right) \left(x - 3\right)$

Example 2

$2 {x}^{2} - 9 x - 5 = \left(2 x + 1\right) \left(x - 5\right)$

Example 3

${x}^{2} - 9 = \left(x + 3\right) \left(x - 3\right)$

I hope that this was helpful.

If both the roots satisfy the quadratic equation, then you have factorised it correctly.

#### Explanation:

I will try explaining with an example.
${x}^{2} + 4 x - 12 = 0$
The roots are $2$ and $- 6$.
Now simply put the values in the quadratic equation.

First, I will try $- 6$
${\left(- 6\right)}^{2} + 4 \left(- 6\right) - 12 = 0$
$36 - 24 - 12 = 0$
$\cancel{36} - \cancel{36} = 0$
$0 = 0$
which is true.

Now, I will try $2$
${2}^{2} + 4 \cdot 2 - 12 = 0$
$4 + 8 - 12 = 0$
$\cancel{12} - \cancel{12} = 0$
$0 = 0$
which is also true.

$\therefore$ Our answer was correct.

Hope you got it :)

• Extension to factoring, when the trinomials do not factor into a square (it also works with squares).

Sum-product-method
Say you have an expression like ${x}^{2} + 15 x + 36$
Then you try to write $36$ as the product of two numbers, and $15$ as the sum (or difference) of the same two numbers. In this case (with both being positive) it's not so hard. You take the sum.

You can write $36 = 1 \cdot 36 = 2 \cdot 18 = 3 \cdot 12 = 4 \cdot 9 = 6 \cdot 6$
Sums of these are $37 , 20 , 15 , 13 , 12$ respectively
Differences are $35 , 16 , 9 , 5 , 0$ respectively
$15 = + 3 + 12$ will do. So the factoring becomes:
$\left(x + 3\right) \left(x + 12\right)$
Check your answer! $= {x}^{2} + 12 x + 3 x + 36$

It's a bit harder when one or two of the numbers are negative, let's take ${x}^{2} - 15 x + 36$
Same as the first, only now both factors are negative
$\left(x - 3\right) \left(x - 12\right) = {x}^{2} - 12 x - 3 x + 36 =$ the original

Extra
If the last number ($36$) is negative, you will have to work with the difference of the factors. Check the next one yourself:
x^2+5x-36=(x+9)(x-4)=?

And now try: x^2-5x-36=?

• Factorization of a quadratic expression is the opposite of expansion, and is the process of putting the brackets back into the expression rather than taking them out.

To factorize a quadratic expression of the form $a {x}^{2} + b x + c$ you must find two numbers that add together to give the first coefficient of $x$ and multiply to give the second coefficient of $x$.

An example of this would be the equation ${x}^{2} + 5 x + 6$, which factorizes to give the expression $\left(x + 6\right) \left(x - 1\right)$

Now, one might expect the solution to include the numbers 2 and 3, as these two numbers both add together to give 5 and multiply to give 6. However, as the signs differ in the factorized equation, then the solution to the equation must be $\left(x + 6\right) \left(x - 1\right)$, as $+ 6 - 1$ gives $5$, and $6 \times 1$ yields a solution of 6.

The equation can be checked by multiplying the solutions back into the equation to give the original quadratic of ${x}^{2} + 5 x + 6$.