# Factorization of a Quadratic Equation

When factorizing quadratic equations, there is generally nothing that is common in every term. In this case, we cannot pull a constant value out and rewrite the problem.

You should know that your answer would look like. (
For example,

Start with the constant term. In this particular case, the constant term is âˆ’12. Factorize âˆ’12 into groups. Each group is a set of two numbers that when multiplied together will give âˆ’12. With the number âˆ’12, these can be the following groups:

You should know that your answer would look like. (

*y*+*a*) Ã— (*y*+*b*). The trick is knowing how to find*a*and*b*.*y*^{2}+ 4*y*âˆ’ 12Start with the constant term. In this particular case, the constant term is âˆ’12. Factorize âˆ’12 into groups. Each group is a set of two numbers that when multiplied together will give âˆ’12. With the number âˆ’12, these can be the following groups:

{âˆ’1,12}, {âˆ’2,6}, (âˆ’3,4}, {1, âˆ’12}, {2, âˆ’6}, {3, âˆ’4}

Now you have six different groups. The next step is to find out the coefficient of the â€˜
If you add âˆ’1 and 12, you get 11. This is not correct, So, try again.
If you add âˆ’2 and 6, then you get 4. This is the required group.

The numbers that you choose in the group (in this case âˆ’2 and 6) will be the

*y*â€™ term and the number to be multiplied by â€˜*y*â€™ in the quadratic equation. In this case, that number is 4. Now, we have to find out which one of the group will give us 4, when the two numbers of each group are added together.The numbers that you choose in the group (in this case âˆ’2 and 6) will be the

*a*and the*b*that we have been looking for. Which one you chose for*a*and which you chose for*b*does not matter. I will just use*a*= âˆ’2 and*b*= 6,[

*y*+ (âˆ’2)] Ã— (*y*+ 6)= (

*y*âˆ’ 2) (*y*+ 6)

# Properties of the Roots

- A polynomial equation of degree
*n*will have*n*roots, real or imaginary. - Complex roots or surds always occur in pair.
- Every equation of an even degree with coefficient of the highest degree term positive and the constant term negative will have at least two real roots. Out of these two real roots, one will be positive and the other one will be odd.
- Any equation with the sum of all of its coefficients equal to zero will have 1 as one of its roots.
*x*^{2}â€“ 5*x*+ 4 = 0 will have 1 as one of its roots. - If all the terms of an equation are +ve and it does not involve any odd powers of
*x*, then this equation will not have any real root.*f*(*x*) =*x*^{4}+*x*^{2}+ 1 = 0 will not have any real root. As*x*^{4}â‰¥ 0,*x*^{2}â‰¥ 0, either of the numbers can become the*a*or*b*, here we will take 1 and 1 is positive. And we know that sum of three positive number. can never be equal to zero for any real value of*x*.

Example-1

How many real roots will be there of the quadratic equation

*f*(*x*) =*x*^{2}+ 5|*x*| + 6 = 0?Solution

If

Or,

So,
But as we have taken

Now, if

Or,

and
Now as we have taken

And
So, no real value of
Alternatively, it is worth observing that all the terms of

*x*> 0, then*f*(*x*) =*x*^{2}+ 5*x*+ 6 = 0Or,

*x*^{2}+ 5*x*+ 6 = (*x*+ 2) (*x*+ 3) = 0So,

*x*= âˆ’2 and*x*= âˆ’3*x*> 0, the negative values of*x*are not admissible.Now, if

*x*< 0, then*f*(*x*) =*x*^{2}âˆ’ 5*x*+ 6 = 0Or,

*x*^{2}âˆ’ 5*x*+ 6 = (*x*âˆ’ 2) (*x*âˆ’ 3) = 0, So,*x*= 2and

*x*= 3*x*< 0, so the positive values of*x*are not admissible.And

*x*= 0 is not possible.*x*can be obtained.*f*(*x*), viz.,*x*^{2}and 5|*x*| and 6 are positive. So, the sum of these three terms cannot be equal to zero for any real value. Hence, no real value of*x*can be obtained.

# Descartesâ€™ Rule of Signs of Roots

Descartesâ€™ rule tells us how to find the maximum number of positive roots in any equation. The maximum number of positive roots in*f*(

*x*) is equal to the change of signs from positive (+ve) to negative (â€“ve) and from negative (â€“ve) to positive (+ve).

With the help of this, we can also find out the maximum number of real roots. To find out the negative roots, we apply the same procedure with

*f*(âˆ’

*x*).

*x*

^{2}+

*x*â€“ 2 = 0

Signs of different terms:

*x*

^{2}has a positive sign,

*x*has a â€˜+â€™ sign, the constant term (2) has a negative sign.

*f*(

*x*) =

*x*

^{2}+

*x*â€“ 2, and their signs are

+ + âˆ’

As there is only one sign change, we can conclude that this equation will have one positive real root.

Now,

*f*(âˆ’*x*) = (âˆ’*x*)^{2}+ (âˆ’*x*) â€“ 2 =*x*^{2}âˆ’*x*â€“ 2Again, there is just one sign change, so there will be one negative root. Also there are no positive or negative imaginary roots and only real numbers can be negative or positive.

Example-2

How

**many real roots will be there for the equation***x*^{4}+*x*^{2}+ 1 = 0?Solution

To find out the positive roots:

*f*(

*x*) =

*x*

^{4}+

*x*

^{2}+ 1, and signs are

+ + +

There is no sign change between any two terms, so, no roots will be positive.

To find out the negative roots:

*f*(âˆ’

*x*) = (âˆ’

*x*)

^{4}+ (âˆ’

*x*)

^{2}+ 1 =

*x*

^{4}+

*x*

^{2}+ 1

Again, there is no sign change, so, the equation will have no negative roots.

And

*x*= 0 is also not a root of this equation.So, no real roots are possible which can satisfy this equation.

It can also be inferred, that all the four roots of this equation will be negative.

Alternatively, we have done this question while seeing the properties of roots (v) also.