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# How to factor $f\left( x \right) = \dfrac{1}{2}{x^2} + \dfrac{5}{2}x - \dfrac{3}{2}$ ?

Last updated date: 02nd Aug 2024
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Hint: Here, we will first take out the common factor from the quadratic equation. Then by using the quadratic roots formula we will find the roots. We will then convert the obtained roots into the factors. And back substitute the factors in the given equation to get the required answer.

Formula Used:
Quadratic roots is given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete Step by Step Solution:
We are given with a Quadratic equation $f\left( x \right) = \dfrac{1}{2}{x^2} + \dfrac{5}{2}x - \dfrac{3}{2}$.
Now, we will take out the common factor from all the terms. Therefore, we get
$\Rightarrow f\left( x \right) = \dfrac{1}{2}\left( {{x^2} + 5x - 3} \right)$ …………………………………………$\left( 1 \right)$
We will now find the factors of the quadratic equation by using the Quadratic roots formula.
Comparing the above equation with the general form of quadratic equation$a{x^2} + bx + c = 0$ , we get
$\begin{array}{l}a = 1\\b = 5\\c = - 3\end{array}$
By substituting the coefficient of ${x^2}$, coefficient of $x$ and the constant term $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, we get
$x = \dfrac{{ - 5 \pm \sqrt {{{\left( 5 \right)}^2} - 4\left( 1 \right)\left( { - 3} \right)} }}{{2\left( 1 \right)}}$
Simplifying the expression, we get
$\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 + 12} }}{2}$
By adding the terms, we get
$\Rightarrow x = \dfrac{{ - 5 \pm \sqrt {37} }}{2}$
Now, by rewriting the equation, we get
$\Rightarrow 2x + 5 = \pm \sqrt {37}$
So, we get two factors from the above equation as:
$\begin{array}{l} \Rightarrow 2x + 5 = + \sqrt {37} \\ \Rightarrow 2x + 5 - \sqrt {37} = 0\end{array}$
And
$\begin{array}{l} \Rightarrow 2x + 5 = - \sqrt {37} \\ \Rightarrow 2x + 5 + \sqrt {37} = 0\end{array}$
Now using these factors we can write the equation $\left( {{x^2} + 5x - 3} \right)$ as:
$\left( {{x^2} + 5x - 3} \right) = \left( {2x + 5 - \sqrt {37} } \right)\left( {2x + 5 + \sqrt {37} } \right)$
When multiplying the factors, we will get the coefficient of ${x^2}$ as $4$ and not $\dfrac{1}{2}$ , so to get the coefficient of ${x^2}$as $\dfrac{1}{2}$, we will divide the product by $\dfrac{1}{8}$.
$\Rightarrow f\left( x \right) = \dfrac{1}{8}\left( {2x + 5 - \sqrt {37} } \right)\left( {2x + 5 + \sqrt {37} } \right)$

Therefore, the factors of $f\left( x \right) = \dfrac{1}{2}{x^2} + \dfrac{5}{2}x - \dfrac{3}{2}$ are $\dfrac{1}{8}\left( {2x + 5 - \sqrt {37} } \right)$and$\left( {2x + 5 + \sqrt {37} } \right)$.

Note:
A quadratic equation is an equation that has the highest degree of 2 and has two solutions. We know that we can solve the quadratic equation by using any of the four methods. Some quadratic equations cannot be solved by using the factorization method and square root method. But here we have used the method of quadratic formula. We should be careful that the quadratic equation should be arranged in the right form. As we have both the positive and negative signs in the formula, so we will write the solutions for the equations according to the signs.