Question

How do you factor the expression ${x^2} + 5x + 4$?

Answer 1

Hint:
We know that for finding the factor and solution of any quadratic equation and polynomial we use the most famous and useful formula which is the Sreedhar Acharya’s formula which is a standard formula for all quadratic equations.

Complete step by step solution:
Given that –
Find the factor and solve the given quadratic polynomial which is ${x^2} + 5x + 4$
Let – the given polynomial as an equation so we can solve it easily ${x^2} + 5x + 4 = 0$
Now we know that the formula of Sreedhar Acharya which is
If the general quadratic equation is the $a{x^2} + bx + c = 0$ then factor and solution of this equation is the
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now we will solve our given equation ${x^2} + 5x + 4 = 0$
Now compare our given equation with the general form of equation $a{x^2} + bx + c = 0$ then
We will get the value of $a,b,c$
Now after comparison the value of $a = 1,b = 5,c = 4$
Now put all value in the formula of Sreedhar Acharya $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ then
Now after putting all values we will get
$ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4 \times 1 \times 4} }}{{2 \times 1}}$
After solving all the calculation, we will get our solution which is
$ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt {{{(25)}^{}} - 16} }}{{2 \times 1}}$
After solving the value of our above equation, we will get
$ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt 9 }}{{2 \times 1}}$
Now we can write it as $x = \dfrac{{ - 5 \pm \sqrt {3 \times 3} }}{2}$
So, we will get two pair of $3$ so we will get $x = \dfrac{{ - 5 \pm 3}}{2}$
So, we got two solution of $x$ one is in the positive form and another is in the form negative so our solution and factor of $x$ is the $x = \dfrac{{ - 5 + 3}}{2} = ( - 1)$ and another is the $x = \dfrac{{ - 5 - 3}}{2} = ( - 4)$

Therefore our quadratic polynomial ${x^2} + 5x + 4$ factor and solutions are $x = ( - 1)$ and $x = ( - 4)$ which is our required answer to our question.

Note:
For solving any quadratic equation quickly we will use the factor method in which we have to find the factor of our given quadratic equation’s midterm (midterm is coefficient of variable $x$ ) such that their addition is equal to midterm and their multiply is equal to constant term, it is very easy and quick method for solving quadratic equation.