Question

How do you write $6{x^2} + 4x$ in factored form?

Answer 1

Hint: First take $6$ common from the given equation. Next, compare the quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Complete step by step answer:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first we will take $6$ common from the given equation.
$ \Rightarrow 6{x^2} + 4x = 6\left( {{x^2} + \dfrac{2}{3}x} \right)$…(i)
Next, compare ${x^2} + \dfrac{2}{3}x = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + \dfrac{2}{3}x = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = \dfrac{2}{3}$ and $c = 0$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( {\dfrac{2}{3}} \right)^2} - 4\left( 1 \right)\left( 0 \right)$
After simplifying the result, we get
$ \Rightarrow D = {\left( {\dfrac{2}{3}} \right)^2}$
Which means given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \dfrac{2}{3} \pm \dfrac{2}{3}}}{{2 \times 1}}$
$ \Rightarrow x = - \dfrac{1}{3} \pm \dfrac{1}{3}$
It can be written as
$ \Rightarrow 3x = 0$ and $x = - \dfrac{2}{3}$
$ \Rightarrow x = 0$ and $x + \dfrac{2}{3} = 0$
Thus, ${x^2} + \dfrac{2}{3}x$ can be factored as $\left( x \right)\left( {x + \dfrac{2}{3}} \right)$.
Now, substitute these factors of ${x^2} + \dfrac{2}{3}x$ in equation (i).
$ \Rightarrow 6{x^2} + 4x = 6\left( x \right)\left( {x + \dfrac{2}{3}} \right)$
It can also be written as
$ \Rightarrow 6{x^2} + 4x = 2x\left( {3x + 2} \right)$

Therefore, $6{x^2} + 4x$ can be factored as $2x\left( {3x + 2} \right)$.

Note: We can directly factor the polynomial by taking $2x$ common from $6{x^2} + 4x$.
Factor $2x$ out of $6{x^2}$.
$2x\left( {3x} \right) + 4x$
Factor $2x$ out of $4x$.
$2x\left( {3x} \right) + 2x\left( 2 \right)$
Take $2x$ common from $2x\left( {3x} \right)$ and $2x\left( 2 \right)$.
$2x\left( {3x + 2} \right)$