Question

How do you factor $ - 12{x^2} + 27$?

Answer 1

Hint: First, equate this polynomial with zero and make it an equation. Next, take $ - 3$ common from the given equation and then divide both sides of the equation by $ - 3$. 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 solution:
First, equate this polynomial with zero and make it an equation.
$ \Rightarrow - 12{x^2} + 27 = 0$
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 $ - 3$ common from the given equation.
$ \Rightarrow - 3\left( {4{x^2} - 9} \right) = 0$
Divide both sides of the equation by $ - 3$.
$ \Rightarrow 4{x^2} - 9 = 0$
Next, compare $4{x^2} - 9 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $4{x^2} - 9 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 4$, $b = 0$ and $c = - 9$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 0 \right)^2} - 4\left( 4 \right)\left( { - 9} \right)$
After simplifying the result, we get
$ \Rightarrow D = 144$
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
$ \Rightarrow x = \dfrac{{ - 0 \pm 12}}{{2 \times 4}}$
Divide numerator and denominator by $4$, we get
$ \Rightarrow x = \pm \dfrac{3}{2}$
$ \Rightarrow 2x = 3$ and $2x = - 3$
$ \Rightarrow 2x - 3 = 0$ and $2x + 3 = 0$
Therefore, the trinomial $4{x^2} - 9$ can be factored as $\left( {2x - 3} \right)\left( {2x + 3} \right)$.
Since, $ - 12{x^2} + 27 = - 3\left( {4{x^2} - 9} \right)$

Therefore, the trinomial $ - 12{x^2} + 27$ can be factored as $ - 3\left( {2x - 3} \right)\left( {2x + 3} \right)$.

Note: We can also factorize a given trinomial using algebraic identity.
Algebraic identity: ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
First take $ - 3$ common from the given expression.
$ - 3\left( {4{x^2} - 9} \right)$
So, rewrite $4{x^2}$ as ${\left( {2x} \right)^2}$.
$ \Rightarrow - 3\left( {{{\left( {2x} \right)}^2} - 9} \right)$
Now, rewrite $9$ as ${3^2}$.
$ \Rightarrow - 3\left( {{{\left( {2x} \right)}^2} - {3^2}} \right)$
Since both terms are perfect squares, factor using the difference of squares formula, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ where $a = 2x$ and $b = 3$.
$ \Rightarrow - 3\left( {2x - 3} \right)\left( {2x + 3} \right)$
Therefore, the trinomial $ - 12{x^2} + 27$ can be factored as $ - 3\left( {2x - 3} \right)\left( {2x + 3} \right)$ or $3\left( {3 - 2x} \right)\left( {3 + 2x} \right)$.