Question

Solve the expression $5 - 4x(1 + 3x)$
$A)(1 + 2x)(5 + 6x)$
$B)(1 - 2x)(5 + 6x)$
$C)(1 - 2x)(5 - 6x)$
$D)(1 + 2x)(5 - 6x)$

Answer 1

Hint: Here we are asked to solve the given quadratic equation. Since it is an equation of order two it will have two roots. The roots of a quadratic equation can be found by using the formula.
The word quadratic means second-degree values of the given variables.
Then by further simplifying, the quadratic equation $a{x^2} + bx + c = 0$ can be written in the form of $(x - \alpha )(x - \beta ) = 0$.
In this equation, the factors of the quadratic equation are $(x - \alpha )$ and $(x - \beta )$.
So, to determine the roots of the equation, these factors of the equation will be made equal to $0$.
That is,
$(x - \alpha ) = 0$ $(x - \beta ) = 0$
Which gives,
$x = \alpha $ $x = \beta $
So, $\alpha $ and $\beta $ are the roots of the quadratic equation.

Complete step-by-step solution:
Since given that $5 - 4x(1 + 3x)$ and by the multiplication operation we get $5 - 4x(1 + 3x) = 5 - 4x - 12{x^2}$
Since given that $ - 12{x^2} - 4x + 5 = 0$. Now convert the value $ - 4x = 6x - 10x$ then we get $ - 12{x^2} + 6x - 10x + 5 = 0$
Now taking the common values out then we have
$ - 12{x^2} + 6x - 10x + 5 = 0 $
$\Rightarrow 6x( - 2x + 1) + 5( - 2x + 1) = 0$
Then by further simplifying, the quadratic equation $a{x^2} + bx + c = 0$ can be written in the form of $(x - \alpha )(x - \beta ) = 0$. Hence, we get
$6x( - 2x + 1) + 5( - 2x + 1) = 0 $
$\Rightarrow (1 - 2x)(6x + 5) = 0$
Therefore $5 - 4x(1 + 3x)$ can be simplified into $(1 - 2x)(6x + 5) = 0$
Thus, the option $B)(1 - 2x)(5 + 6x)$ is correct.

Note: The quadratic equation $a{x^2} + bx + c = 0$ can be factored in the form of $(x - \alpha )(x - \beta ) = 0$ where $(x - \alpha )$ and $(x - \beta )$ are the factors of the quadratic equation so $\alpha $ and $\beta $ are the roots of the quadratic equation and the values of $a$ cannot be $0$.
If the value of $a$ is $0$, then the quadratic equation will become a binomial equation.
Using the multiplication operation we found the values $5 - 4x(1 + 3x) = 5 - 4x - 12{x^2}$
We are also able to make use of the formula of the quadratic equation. Let \[a{x^2} + bx + c = 0\] be a quadratic equation then the roots of this equation are given by \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].