Question

How do you factor $y = 4{x^2} - 4x - 5$?

Answer 1

Hint: First take $4$ common from the given equation. Next, compare the given 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.

Formula used:
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 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

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 $4$ common from the given equation.
$ \Rightarrow y = 4\left( {{x^2} - x - \dfrac{5}{4}} \right)$…(i)
Next, compare ${x^2} - x - \dfrac{5}{4} = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - x - \dfrac{5}{4} = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 1$ and $c = - \dfrac{5}{4}$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 1} \right)^2} - 4\left( 1 \right)\left( { - \dfrac{5}{4}} \right)$
After simplifying the result, we get
$ \Rightarrow D = 1 + 5$
$ \Rightarrow D = 6$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt 6 }}{{2 \times 1}}$
$ \Rightarrow x = \dfrac{1}{2} \pm \dfrac{{\sqrt 6 }}{2}$
It can be written as
$ \Rightarrow x = \dfrac{1}{2} + \dfrac{{\sqrt 6 }}{2}$ and $x = \dfrac{1}{2} - \dfrac{{\sqrt 6 }}{2}$
$ \Rightarrow x - \dfrac{1}{2} - \dfrac{{\sqrt 6 }}{2} = 0$ and $x - \dfrac{1}{2} + \dfrac{{\sqrt 6 }}{2} = 0$
Thus, ${x^2} - x - \dfrac{5}{4}$ can be factored as $\left( {x - \dfrac{1}{2} - \dfrac{{\sqrt 6 }}{2}} \right)\left( {x - \dfrac{1}{2} + \dfrac{{\sqrt 6 }}{2}} \right)$.
Now, substitute these factors of ${x^2} - x - \dfrac{5}{4}$ in equation (i).
$ \Rightarrow y = 4\left( {x - \dfrac{1}{2} - \dfrac{{\sqrt 6 }}{2}} \right)\left( {x - \dfrac{1}{2} + \dfrac{{\sqrt 6 }}{2}} \right)$

Therefore, $y = 4{x^2} - 4x - 5$ can be factored as $y = 4\left( {x - \dfrac{1}{2} - \dfrac{{\sqrt 6 }}{2}} \right)\left( {x - \dfrac{1}{2} + \dfrac{{\sqrt 6 }}{2}} \right)$.

Note: In above question, it should be noted that we get $x = \dfrac{1}{2} + \dfrac{{\sqrt 6 }}{2}$ and $x = \dfrac{1}{2} - \dfrac{{\sqrt 6 }}{2}$ as the roots of equation ${x^2} - x - \dfrac{5}{4} = 0$. No other roots will satisfy the condition. If we take wrong factors, then we will not get a trinomial on their product. So, carefully find the roots.