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# Solve x^2+5x+6=0 by factorizing?

The given equation is a quadratic equation which means the power of the variable is 2. It is also known as a polynomial of the second degree.

It is usually written in the form of ax^2 + bx + c =0. Here a, b and c belong to a set of real numbers and where a is never equal to 0 (a ≠ 0).

For all quadratic equations, there are two values of the variables that satisfy the equation and turn it zero. These are called the roots of the equation. There are two main ways to find the roots of the equation.

These are

• Performing factorization
• Using quadratic formula to find roots.

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Let’s discuss the steps used to solve a quadratic equation.

Steps to factorize a given equation

When we say factorize a given quadratic equation, it means breaking the equation into a product of two factors.

In a nutshell, let’s understand the steps that are followed while factoring a quadratic equation.

• Expand the expression if needed.
• Take all terms to the left-hand side of the equal to sign so the right side becomes zero.
• Arrange all like terms together. Resolve fractions (if any).
• Break the middle term i.e. the coefficient of the variable with power 1.

In ax^2 + bx + c = 0, we need to break b into two parts in such a way that on multiplying them you get ac and on adding you get b.

• Now you have the two factors. Now separate them and equate each of them to zero.
• You are left with two linear equations that are easy to solve.

So, let’s start solving the equation x^+ 5x + 6 = 0

Here, a  = 1, b = 5 and c = 6;

ac = 6 and b = 5

Here, we need to break 5 into two parts in such a way that it gives 6 ( a*c = 1* 6 = 6 ) on multiplication and the addition should be 5 (b).

So, the two values are 2 and 3 which in addition give 5 and on multiplication give 6.

x^+ 5x + 6  can be written as x^2 +  3x + 2x + 6

Here, comes the splitting part. Taking out x (common) thing from the first two terms and taking 2 common factors from the last two terms.

x^+ 5x + 6 = x ( x + 3) + 2 (x + 3)

Taking aside (x + 3), we get two linear equations.

It becomes:

(x + 2) (x + 3) = 0

Now we have got two linear equations (x + 2) and (x + 3).

Equate them to zero individually.

x + 2 = 0 and x + 3 = 0

This makes, x = -2 and x = -3

So, the roots or zeros of x^+ 5x + 6 equation are -2 and -3.

Another method is by using a quadratic formula:

The roots of any equation can be solved by using the following formula

$x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$

First root,  $x=\frac{{-b+\sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$ and the second root is  $x=\frac{{-b-\sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$

By putting the values of a, b and c

We get, the first root as

$x=\frac{{-5~+~\sqrt[{}]{{{{5}^{2}}-4*6}}}}{{2*1}}$

x =( -5 + √1)  / 2

x = -4 / 2

x =  -2

We get the second root as:

$x=\frac{{-5~-~\sqrt[{}]{{{{5}^{2}}-4*6}}}}{{2*1}}$

x =( -5 – √1)  / 2

x = -6 / 2

x =  -3

So, by using the quadratic formula we get the roots of x^+ 5x + 6 as -2 and -3.

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